Class 10 NCERT Introduction to Trigonometry

NCERT

Class 10 NCERT

1.   1- In $\Delta$ABC right angled at B, AB = 24 cm, BC = 7 m. Determine$\\$ (i) sin A, cos A$\\$ (ii) sin C, cos C

Applying Pythagoras theorem for $\Delta$ABC, we obtain$\\$ $AC^2 = AB^2 + BC^2$$\\$= $(24 cm)^2 + (7 cm)^2$$\\$ = $(576 + 49) cm^2$$\\$ = $625 cm^2$$\\$$\therefore AC = \sqrt{625} cm = 25 cm$$\\$

None

$(i) sin A = \dfrac{side opposite to \angle A}{Hypotenuse} = \dfrac{BC}{AC}$$\\$$=\dfrac{7}{25}$$\\$$(ii) cos A = \dfrac{side adjacent to \angle A}{Hypotenuse} = \dfrac{AB}{AC} = \dfrac{24}{25}$$\\$ figure

2.   In Fig. $8.13$, find $tan\ P - cot\ R.$ $\\$ DIAGRAM

Applying Pythagoras theorem for $\Delta{PQR}$, we obtain $\\$ $PR^2 = PQ^2 + QR^2$ $\\$ $(13\ cm)^2 = (12\ cm)^2 + QR^2$ $\\$ $169\ cm^2 = 144\ cm^2 + QR^2$ $\\$ $25\ cm^2 = QR^2$ $\\$ $QR = 5\ cm$

diagram

$tan\ P = \dfrac{Side\ opposite\ to \angle{P}}{Side\ adjacent\ to\ \angle{P}} = \dfrac{QR}{PQ}$ $\\$ $= \dfrac{5}{12}$ $\\$ $cot\ R = \dfrac{Side\ adjacent\ to\ \angle{R}}{Side\ opposite\ to \angle{R}} = \dfrac{QR}{PQ}$ $\\$ $= \dfrac{5}{12}$ $\\$ $tan\ P - cot\ R = \dfrac{5}{12} - \dfrac{5}{12} = 0$

3.   If $sin\ A = \dfrac{3}{4}$, calculate $cos\ A$ and $tan\ A$

Let $\Delta{ABC}$ be a right-angled triangle, right-angled at point $B$. $\\$

diagram

Given that, $\\$ $sin\ A = \dfrac{3}{4}$ $\\$ $\dfrac{BC}{AC} = \dfrac{3}{4}$ $\\$ Let BC be 3k. Therefore, AC will be 4k, where k is a positive integer. Applying Pythagoras theorem in $\Delta{ABC}$, we obtain $\\$ $AC^2 = AB^2 + BC^2$ $\\$ $(4k)^2 = AB2 + (3k)^2$ $\\$ $16k^2 - 9k^2 = AB^2$ $\\$ $7k^2 = AB^2$ $\\$ $AB = \sqrt{7}k$ $\\$ $cos\ A = \dfrac{Side\ adjacent\ to\ \angle{A}}{Hypotenuse}$ $\\$ $= \dfrac{AB}{AC} = \dfrac{\sqrt{7}k}{4k} = \dfrac{\sqrt{7}}{4}$ $\\$ $tan\ A = \dfrac{Side\ opposite\ to\ \angle{A}}{Side\ adjacent\ to\ \angle{A}}$ $\\$ $= \dfrac{BC}{AB} = \dfrac{3k}{\sqrt{7}k} = \dfrac{3}{\sqrt{7}}$ $\\$

4.   Given $15\ cot\ A = 8$, find $sin\ A$ and $sec\ A.$

Consider a right-angled triangle, right-angled at $B.$ $\\$

diagram

$cot\ A = \dfrac{Side\ adjacent\ to \angle{A}}{Side\ opposite\ to\ \angle{A}}$ $\\$ $= \dfrac{AB}{BC}$ $\\$ It is given that, $cot\ A = \dfrac{8}{15}$ $\\$ $\dfrac{AB}{BC} = \dfrac{8}{15}$ $\\$ Let $AB$ be $8k$.Therefore, $BC$ will be $15k$, where k is a positive integer. $\\$ Applying Pythagoras theorem in $\Delta{ABC}$, we obtain $\\$ $AC^2 = AB^2 + BC^2$ $\\$ $= (8k)^2 + (15k)^2$ $\\$ $= 64k^2 + 225k^2$ $\\$ $= 289k^2$ $\\$ $AC = 17k$ $\\$ $sin\ A = \dfrac{Side\ opposite\ to\ \angle{A}}{Hypotenuse} = \dfrac{BC}{AC}$ $\\$ $= \dfrac{15k}{17k} = \dfrac{15}{17}$ $\\$ $sec\ A = \dfrac{Hypotenuse}{Side\ adjacent\ to\ \angle{A}}$ $\\$ $\dfrac{AC}{AB} = \dfrac{17}{18}$ $\\$

5.   Given $sec\ \theta = \dfrac{13}{12}$, calculate all other trigonometric ratios.

Consider a right-angle triangle $\Delta{ABC}$, right-angled at point $B.$

diagram

$sec\ \theta = \dfrac{Hypotenuse}{Side\ adjacent\ to\ \angle{\theta}}$ $\\$ $\dfrac{13}{12}$ $\\$ If $AC$ is $13k,\ AB$ will be $12k$, where $k$ is a positive integer. $\\$ Applying Pythagoras theorem in $\Delta{ABC}$, we obtain $\\$ $(AC)^2 = (AB)^2 + (BC)^2$ $\\$ $(13k)^2 = (12k)^2 + (BC)^2$ $\\$ $169k^2 = 144k^2 + BC^2$ $\\$ $25k^2 = BC^2$ $\\$ $BC = 5k$ $\\$ $sin \theta = \dfrac{Side\ opposite\ to\ \angle{\theta}}{Hypotenuse} = \dfrac{BC}{AC} = \dfrac{5k}{13k} = \dfrac{5}{13}$ $\\$ $cos \theta = \dfrac{Side\ adjacent\ to\ \angle{\theta}}{Hypotenuse} = \dfrac{AB}{AC} = \dfrac{12k}{13k} = \dfrac{12}{13}$ $\\$ $tan \theta = \dfrac{Side\ opposite\ to\ \angle{\theta}}{Side\ adjacent\ to\ \angle{\theta}} = \dfrac{BC}{AB} = \dfrac{5k}{12k} = \dfrac{5}{12}$ $\\$ $cot \theta = \dfrac{Side\ adjacent\ to\ \angle{\theta}}{Side\ opposite\ to\ \angle{\theta}} = \dfrac{AB}{BC} = \dfrac{12k}{5k} = \dfrac{12}{5}$ $\\$ $cosec \theta = \dfrac{Hypotenuse}{Side\ opposite\ to\ \angle{\theta}} = \dfrac{AC}{BC} = \dfrac{13k}{5k} = \dfrac{13}{5}$ $\\$

6.   If $\angle{A}$ and $\angle{B}$ are acute angles such that $cos\ A = cos\ B$, then show that $\angle{A} = \angle{B}.$

Let us consider a triangle $ABC$ in which $CD \bot AB.$

It is given that $\\$ $cos\ A = cos\ B$ $\\$ $\Rightarrow \dfrac{AD}{AC} = \dfrac{BD}{BC} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$ $\\$ We have to prove $\angle{A} = \angle{B}$. To prove this, let us extend $AC$ to $P$ such that $BC = CP.$ $\\$ From equation (1), we obtain $\\$ $\dfrac{AD}{BD} = \dfrac{AC}{BC}$ $\\$ $\Rightarrow \dfrac{AD}{BD} = \dfrac{AC}{CP} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $ (By construction, we have $BC = CP$) $ \ \ \ \ \ \ \ \ \ \ ....(2)$

By using the converse of $B.P.T$, $\\$ $CD | | BP$ $\\$ $\Rightarrow \angle{ACD} = \angle{CPB} $ (Corresponding angles) $ \ \ \ … (3)$ $\\$ And, $\angle{BCD} = \angle{CBP}$ (Alternate interior angles) $ \ \ \ \ \ … (4)$ $\\$ By construction, we have $BC = CP$.$\\$ $\therefore \angle{CBP} = \angle{CPB}$ (Angle opposite to equal sides of a triangle) $ \ \ \ \ \ \ … (5)$ $\\$ From equations $(3),\ (4),$ and $(5)$, we obtain $\\$ $\angle{ACD} = \angle{BCD} \ \ \ \ \ … (6)$ $\\$ In $\Delta{CAD}$ and $\Delta{CBD}$, $\\$ $\angle{ACD} = \angle{BCD}$ [Using equation $(6)$] $\\$ $\angle{CDA} = \angle{CDB}$ [Both $90°$] $\\$ Therefore, the remaining angles should be equal. $\\$ $\therefore \angle{CAD} = \angle{CBD} $ $\\$ $\Rightarrow \angle{A} = \angle{B}$ $\\$ Alternatively, $\\$

Let us consider a triangle $ABC$ in which $CD \bot AB$. $\\$ It is given that, $\\$ $cos\ A = cos\ B$ $\\$ $\Rightarrow \dfrac{AD}{AC} = \dfrac{BD}{BC}$ $\\$ $\Rightarrow \dfrac{AD}{BD} = \dfrac{AC}{BC}$ $\\$ Let $\dfrac{AD}{BD} = \dfrac{AC}{BC} = k$ $\Rightarrow AD = k\ BD \ \ \ \ \ .... (1)$ $\\$ And, $AC = k\ BC \ \ \ \ … (2)$ $\\$ $\Rightarrow (k\ BC)^2 - (k\ BD)^2 = BC^2 - BD^2$ $\\$ $\Rightarrow k^2 (BC^2 - BD^2) = BC^2 - BD^2$ $\\$ $\Rightarrow k^2 = 1$ $\\$ $\Rightarrow k = 1$ $\\$ Putting this value in equation (2), we obtain $\\$ $AC = BC$ $\\$ $\Rightarrow \angle{A} = \angle{B}$ (Angles opposite to equal sides of a triangle)$

7.   If $cot\ \theta= \dfrac{7}{8}$ evaluate : $\\$ (i) $\dfrac{(1 + sin\ \theta)(1 + cos\ \theta)}{(1 - sin\ \theta)(1 - cos\ \theta)}$, $\\$ (ii) $cot^2\ \theta$

Let us consider a right triangle $ABC$, right-angled at point $B.$ $\\$

diagram

$cot\ \theta = \dfrac{Side\ adjacent\ to\ \angle{\theta}}{Side\ opposite\ to\ \angle{\theta}} = \dfrac{BC}{AB}$ $\\$ $= \dfrac{7}{8}$ $\\$ If $BC$ is $7k$, then $AB$ will be $8k$, where $k$ is a positive integer. $\\$ Applying Pythagoras theorem in $\Delta{ABC}$, we obtain $\\$ $AC^2 = AB^2 + BC^2$ $\\$ $= (8k)^2 + (7k)^2$ $\\$ $= 64k^2 + 49k^2$ $\\$ $= 113k^2$ $\\$ $AC = \sqrt{113}k$ $\\$

$sin\ \theta = \dfrac{Side\ opposite\ to\ \angle{\theta}}{Hypotenuse} = \dfrac{AB}{AC}$ $\\$ $= \dfrac{8k}{\sqrt{113}k} = \dfrac{8}{\sqrt{113}}$ $\\$ $cos\ \theta = \dfrac{Side\ adjacent\ to\ \angle{\theta}}{Hypotenuse} = \dfrac{BC}{AC}$ $\\$ $= \dfrac{7k}{\sqrt{113}k} = \dfrac{7}{\sqrt{113}}$ $\\$ $\dfrac{(1 + sin\ \theta)(1 - sin\ \theta)}{(1 + cos\ \theta)(1 - cos\ \theta)} = \dfrac{(1 - sin^2\ \theta)}{(1 - cos^2\ \theta)}$ $\\$ (i) $=\dfrac{1 - \Big(\dfrac{8}{\sqrt{113}}\Big)^2}{1 - \Big(\dfrac{7}{\sqrt{113}}\Big)} = \dfrac{1 - \dfrac{64}{113}}{1 - \dfrac{49}{113}}$ $\\$ $= \dfrac{\dfrac{49}{113}}{\dfrac{64}{113}} = \dfrac{49}{64}$ $\\$ (ii) $cot^2 \theta = (cot\ \theta)^2 = \Big(\dfrac{7}{8}\Big)^2 = \dfrac{49}{64}$ $\\$

8.   If $3\ cot\ A = 4$, check whether $\dfrac{1 - tan^2\ A}{1 + tan^2\ A} = cos^2\ A - sin^2\ A$ or not.

It is given that $3cot\ A = 4$ $\\$ Or, $cot\ A = \dfrac{4}{3}$ $\\$ Consider a right triangle $ABC$, right-angled at point $B$. $\\$ $cot\ A = \dfrac{Side\ adjacent\ to\ \angle{A}}{Side\ opposite\ to\ \angle{A}}$ $\\$ $\dfrac{AB}{BC} = \dfrac{4}{3}$ $\\$ If $AB$ is $4k$, then $BC$ will be $3k$, where $k$ is a positive integer. $\\$ In $\Delta{ABC}$, $\\$ $(AC)^2 = (AB)^2 + (BC)^2$ $\\$ $= (4k)^2 + (3k)^2$ $\\$ $= 16k^2 + 9k^2$ $\\$ $= 25k^2$ $\\$ $AC = 5k$ $\\$ $cos A = \dfrac{Side\ adjacent\ to \angle{A}}{Hypotenuse} = \dfrac{AB}{AC}$ $\\$ $= \dfrac{4k}{5k} = \dfrac{4}{5}$ $\\$ $sin\ A = \dfrac{Side\ opposite\ to\ \angle{A}}{Hypotenuse} = \dfrac{BC}{AC}$ $\\$ $tan A = \dfrac{Side\ opposite\ to\ \angle{A}}{Hypotenuse} = \dfrac{BC}{AB}$ $\\$ $= \dfrac{3k}{4k} = \dfrac{3}{4}$ $\\$ $ \dfrac{1 - tan^2\ A}{1 + tan^2\ A} = \dfrac{1 - \Big(\dfrac{3}{4}\Big)^2}{ 1 + \Big(\dfrac{3}{4}\Big)^2} = \dfrac{1 - \dfrac{9}{16}}{1 + \dfrac{9}{16}}$ $\\$ $= \dfrac{\dfrac{7}{16}}{\dfrac{25}{16}} = \dfrac{7}{25}$ $\\$ $cos^2\ A - sin^2\ A = \Big(\dfrac{4}{5}\Big)^2 - \Big(\dfrac{3}{5}\Big)^2$ $\\$ $= \dfrac{16}{25} - \dfrac{9}{25} = \dfrac{7}{25}$ $\\$ $\therefore \dfrac{1 - tan^2 A}{1 + tan^2 A} = cos^2\ A - sin^2\ A$

9.   In triangle $ABC$, right-angled at $B$, if $tan\ A = \dfrac{1}{\sqrt{3}}$, Find the value of : $\\$ (i) $sin\ A\ cos\ C + cos\ A\ sin\ C$ $\\$ (ii) $cos\ A\ cos\ C - sin\ A\ sin\ C$

$tan A = \dfrac{1}{\sqrt{3}}$ $\\$ $\dfrac{BC}{AB} = \dfrac{1}{\sqrt{3}}$ $\\$ If $BC$ is $k$, then $AB$ will be , where $k$ is a positive integer. $\\$ In $\Delta{ABC},$ $\\$ $AC^2 = AB^2 + BC^2$ $\\$ $ = (\sqrt{3}k)^2 + (k)^2$ $\\$ $= 3k^2 + k^2 = 4k^2$ $\\$ $\therefore AC = 2k$ $\\$

$sin\ A = \dfrac{Side\ opposite\ to\ \angle{A}}{Hypotenuse} = \dfrac{BC}{AC} = \dfrac{k}{2k} = \dfrac{1}{2}$ $\\$ $cos\ A = \dfrac{Side\ adjacent\ to\ \angle{A}}{Hypotenuse} = \dfrac{AB}{AC} = \dfrac{\sqrt{3}k}{2k} = \dfrac{\sqrt{3}}{2}$ $\\$ $sin\ C = \dfrac{Side\ opposite\ to\ \angle{C}}{Hypotenuse} = \dfrac{AB}{AC} = \dfrac{\sqrt{3}k}{2k} = \dfrac{\sqrt{3}}{2}$ $\\$ $cos\ C = \dfrac{Side\ adjacent\ to\ \angle{C}}{Hypotenuse} = \dfrac{BC}{AC} = \dfrac{k}{2k} = \dfrac{1}{2}$ $\\$ (i) $sin\ A\ cos\ C + cos\ A\ sin\ C$ $\\$ $= \Big(\dfrac{1}{2}\Big) \Big(\dfrac{1}{2}\Big) + \Big(\dfrac{\sqrt{3}}{2}\Big) \Big(\dfrac{\sqrt{3}}{2}\Big) = \dfrac{1}{4} + \dfrac{3}{4}$ $\\$ $= \dfrac{4}{4} = 1$ $\\$ (ii) $cos\ A\ cos\ C - sin\ A\ sin\ C$ $\\$ $= \Big(\dfrac{\sqrt{3}}{2}\Big) \Big(\dfrac{1}{2}\Big) - \Big(\dfrac{1}{2}\Big) \Big(\dfrac{\sqrt{3}}{2}\Big) = \dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4} = 0$ $\\$

10.   In $\Delta{PQR}$, right-angled at $Q,\ PR + QR = 25\ cm$ and $PQ = 5\ cm$. Determine the values of $sin\ P,\ cos\ P$ and $tan\ P$

Given that, $PR + QR = 25$ $\\$ $PQ = 5$ $\\$ Let $PR$ be $x.$ $\\$ Therefore, $QR = 25 - x$ $\\$ Applying Pythagoras theorem in $\Delta{PQR}$, we obtain $\\$ $PR^2 = PQ^2 + QR^2$ $\\$ $x^2 = (5)^2 + (25 - x)^2$ $\\$ $x^2 = 25 + 625 + x^2 - 50x$ $\\$ $50x = 650$ $\\$ $x = 13$ $\\$ Therefore, $PR = 13\ cm$ $\\$ $QR = (25 - 13)\ cm = 12\ cm$ $\\$ $sin\ P = \dfrac{Side\ opposite\ to\ \angle{P}}{Hypotenuse} = \dfrac{QR}{PR} = \dfrac{12}{13}$ $\\$ $cos\ P = \dfrac{Side\ adjacent\ to\ \angle{P}}{Hypotenuse} = \dfrac{PQ}{PR} = \dfrac{5}{13}$ $\\$ $tan\ P = \dfrac{Side\ opposite\ to\ \angle{P}}{Hypotenuse} = \dfrac{QR}{PR} = \dfrac{12}{5}$ $\\$

11.   State whether the following are true or false. Justify your answer. $\\$ (i) The value of $tan\ A$ is always less than $1$. $\\$ (ii) $sec\ A = {12}{5}$ for some value of angle $A$. $\\$ (iii) $cos\ A$ is the abbreviation used for the cosecant of angle $A$. $\\$ (iv) $cot\ A$ is the product of $cot$ and $A$. $\\$ (v) $sin\ \theta= \dfrac{4}{3}$ for some angle $\theta$.

(i) Consider a $\Delta{ABC}$, right-angled at $B$. $\\$

diagram

$tan\ A = \dfrac{Side\ opposite\ to\ \angle{A}}{Side\ adjacent\ to\ \angle{A}}$ $\\$ $= \dfrac{12}{5}$ $\\$ But $\dfrac{12}{5} > 1$ $\\$ $\therefore tan\ A > 1$ $\\$ So, $tan\ A < 1$ is not always true.$\\$ Hence, the given statement is false.$\\$ (ii) $sec\ A = \dfrac{12}{5}$ $\\$

diagram

$\dfrac{Hypotenuse}{Side\ adjacent\ to\ \angle{A}} = \dfrac{12}{5}$ $\\$ $\dfrac{AC}{AB} = \dfrac{12}{5}$ $\\$ Let $AC$ be $12k$, $AB$ will be $5k$, where $k$ is a positive integer. $\\$ Applying Pythagoras theorem in $\Delta{ABC}$, we obtain $\\$ $AC^2 = AB^2 + BC^2$ $\\$ $(12k)^2 = (5k)^2 + BC^2$ $\\$ $144k^2 = 25k^2 + BC^2$ $\\$ $BC^2 = 119k^2$ $\\$ $BC = 10.9k$ $\\$ It can be observed that for given two sides $AC = 12k$ and $AB = 5k$, $\\$ BC should be such that, $\\$ $AC - AB < BC < AC + AB$ $\\$ $12k - 5k < BC < 12k + 5k$ $\\$ $7k < BC < 17\ k$ $\\$ However, $BC = 10.9k$. Clearly, such a triangle is possible and hence, such value of $sec\ A$ is possible. $\\$ Hence, the given statement is true. $\\$

(iii) Abbreviation used for cosecant of angle $A$ is $cosec\ A$. And $cos\ A$ is the abbreviation used for cosine of angle $A$. $\\$ Hence, the given statement is false. $\\$

(iv) $cot\ A$ is not the product of $cot$ and $A$. It is the cotangent of $\angle{A}$. Hence, the given statement is false.

(v) $sin\ \theta = \dfrac{4}{3}$ $\\$ We know that in a right-angled triangle, $\\$ $sin\ \theta = \dfrac{Side\ opposite\ to\ \angle{\theta}}{Hypotenuse}$ $\\$ In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of $sin\ \theta$ is not possible. Hence, the given statement is false

12.   Evaluate the following : $\\$ (i) $sin\ 60°\ cos\ 30° + sin\ 30°\ cos\ 60°$ $\\$ (ii) $2\ tan^2\ 45° + cos^2\ 30° - sin^2\ 60°$ $\\$ (iii) $\dfrac{cos\ 45°}{sec\ 30° + cosec\ 30°}$ $\\$ (iv) $\dfrac{sin\ 30° + tan\ 45° - cosec\ 60°}{sec\ 30° + cos\ 60° + cot\ 45°}$ $\\$ (v) $\dfrac{5\ cos^2\ 60^o + 4\ sec^2\ 30^o - tan^2\ 45^o}{sin^2\ 30^o + cos^2\ 30^o}$

(i) $sin\ 60°\ cos\ 30° + sin\ 30°\ cos\ 60°$ $\\$ $\Big(\dfrac{\sqrt{3}}{2}\Big)\Big(\dfrac{\sqrt{3}}{2}\Big) + \Big(\dfrac{1}{2}\Big)\Big(\dfrac{1}{2}\Big)$ $\\$ $ = \dfrac{3}{4} + \dfrac{1}{4} = \dfrac{4}{4} = 1$ $\\$ (ii) $2\ tan^2\ 45° + cos^2\ 30° - sin^2\ 60°$ $\\$ $= 2(1)^2 + \Big(\dfrac{\sqrt{3}}{2}\Big)^2 - \Big(\dfrac{\sqrt{3}}{2}\Big)^2 $ $\\$ $= 2 + \dfrac{3}{4} - \dfrac{3}{4} = 2$ $\\$

(iii) $\dfrac{cos\ 45°}{sec\ 30° + cosec\ 30°}$ $\\$ $= \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}} + 2} = \dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2 + 2\sqrt{3}}{\sqrt{3}}}$ $\\$ $= \dfrac{\sqrt{3}}{\sqrt{2}(2 + 2\sqrt{3})} = \dfrac{\sqrt{3}}{2\sqrt{2} + 2\sqrt{6}}$ $\\$ $ = \dfrac{\sqrt{3}(2\sqrt{6} - 2\sqrt{2})}{(2\sqrt{6} + 2\sqrt{2})(2\sqrt{6} - 2\sqrt{2})}$ $\\$ $= \dfrac{2\sqrt{3}(\sqrt{6} - \sqrt{2})}{(2\sqrt{6})^2 - (2\sqrt{2})^2} = \dfrac{2\sqrt{3}(\sqrt{6} - \sqrt{2})}{24 - 8} = \dfrac{2\sqrt{3}(\sqrt{6} - \sqrt{2})}{16}$ $\\$ $= \dfrac{\sqrt{18} - \sqrt{6}}{8} = \dfrac{3\sqrt{2} - \sqrt{6}}{8}$

(iv) $\dfrac{sin\ 30° + tan\ 45° - cosec\ 60°}{sec\ 30° + cos\ 60° + cot\ 45°}$ $\\$ $= \dfrac{\dfrac{1}{2} + 1 - \dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}} + \dfrac{1}{2} + 1} = \dfrac{\dfrac{3}{2} - \dfrac{2}{\sqrt{3}}}{\dfrac{3}{2} + \dfrac{2}{\sqrt{3}}}$ $\\$ $= \dfrac{\dfrac{3\sqrt{3} - 4}{2\sqrt{3}}}{\dfrac{3\sqrt{3} + 4}{2\sqrt{3}}} = \dfrac{(3\sqrt{3} - 4)}{(3\sqrt{3} + 4)}$ $\\$ $= \dfrac{(3\sqrt{3} - 4)(3\sqrt{3} - 4)}{(3\sqrt{3} + 4)(3\sqrt{3} - 4)} = \dfrac{(3\sqrt{3} - 4)^2}{(3\sqrt{3})^2 - (4)^2}$ $\\$ $= \dfrac{27 + 16 - 24\sqrt{3}}{27 - 16} = \dfrac{43 - 24\sqrt{3}}{11}$

(v) $\dfrac{5\ cos^2\ 60^o + 4\ sec^2\ 30^o - tan^2\ 45^o}{sin^2\ 30^o + cos^2\ 30^o}$ $\\$ $\dfrac{5 \Big(\dfrac{1}{2}\Big)^2 + 4 \Big(\dfrac{2}{\sqrt{3}}\Big)^2 - (1)^2}{\Big(\dfrac{1}{2}\Big)^2 + (\dfrac{\sqrt{3}}{2})^2}$ $\\$ $\dfrac{5 \Big(\dfrac{1}{4}\Big) + 4 \Big(\dfrac{16}{3}\Big) -1}{\dfrac{1}{4} + \dfrac{3}{4}}$ $\\$ $= \dfrac{\dfrac{15 + 64 -12}{12}}{\dfrac{4}{4}} = \dfrac{67}{12}$ $\\$

13.   Choose the correct option and justify your choice : $\\$ (i) $\dfrac{2\ tan\ 30^o}{1 + tan^2\ 30^o} =$ $\\$ $(A)\ sin\ 60° \ \ \ \ \ \ \ \ (B)\ cos\ 60° \ \ \ \ \ \ \ \ (C)\ tan\ 60° \ \ \ \ \ \ \ \ (D)\ sin\ 30°$ $\\$ (ii) $\dfrac{1 - tan^2\ 45^o}{1 + tan^2\ 45^o}$ $\\$ $(A)\ tan\ 90° \ \ \ \ \ \ \ \ (B)\ 1 \ \ \ \ \ \ \ \ (C)\ sin\ 45° \ \ \ \ \ \ \ \ (D)\ 0$ $\\$ (iii) $sin\ 2A = 2\ sin\ A$ is true when $A =$ $\\$ $(A)\ 0° \ \ \ \ \ \ \ \ (B)\ 30° \ \ \ \ \ \ \ \ (C)\ 45° \ \ \ \ \ \ \ \ (D)\ 60°$ $\\$ (iv) $\dfrac{2\ tan\ 30^o}{1 - tan^2\ 30^o} =$ $\\$ $(A)\ cos\ 60° \ \ \ \ \ \ \ \ (B)\ sin\ 60° \ \ \ \ \ \ \ \ (C)\ tan\ 60° \ \ \ \ \ \ \ \ (D)\ sin\ 30°$

Answer

13   None

(i) $\dfrac{2\ tan\ 30^o}{1 + tan^2\ 30^o} =$ $\\$

$= \dfrac{2\Big(\dfrac{1}{\sqrt{3}}\Big)}{1 + \Big(\dfrac{1}{\sqrt{3}}\Big)^2} = \dfrac{\dfrac{2}{\sqrt{3}}}{1 + \dfrac{1}{3}} = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{4}{3}}$ $\\$ $ = \dfrac{6}{4\sqrt{3}} = \dfrac{\sqrt{3}}{2}$ $\\$ Out of the given alternatives, only $sin\ 60^o = \dfrac{\sqrt{3}}{2}$ $\\$ Hence, $(A)$ is correct.

(ii) $\dfrac{1 - tan^2\ 45^o}{1 + tan^2\ 45^o}$ $\\$ $= \dfrac{1 - (1)^2}{1 + (1)^2} = \dfrac{1 - 1}{1 + 1} = \dfrac{0}{2} = 0$ $\\$ Hence, $(D)$ is correct.

(iii) $sin\ 2A = 2\ sin\ A$ is true when $A =$ $\\$ Out of the given alternatives, only $A = 0°$ is correct. $\\$ As $sin\ 2A = sin\ 0° = 0$ $\\$ $2\ sinA = 2sin\ 0° = 2(0) = 0$ $\\$ Hence, $(A)$ is correct.

(iv) $\dfrac{2\ tan\ 30^o}{1 - tan^2\ 30^o} =$ $\\$ $ = \dfrac{2\Big(\dfrac{1}{\sqrt{3}}\Big)}{1 - \Big(\dfrac{1}{\sqrt{3}}\Big)^2} = \dfrac{\dfrac{2}{\sqrt{3}}}{1 - \dfrac{1}{3}} = \dfrac{\dfrac{2}{\sqrt{3}}}{\dfrac{2}{3}}$ $\\$ $= \sqrt{3}$ $\\$ Out of the given alternatives, only tan $60° = \sqrt{3}$ $\\$ Hence, $(C)$ is correct.

14.   If tan $(A + B) = \sqrt{3}$ and $tan\ (A – B) = \dfrac{1}{\sqrt{3}} ; 0^o < A + B \le 90^o ; A > B,$ Find $A$ and $B$.

$tan\ (A + B) = \sqrt{3}$ $\\$ $\Rightarrow tan\ (A + B) = tan\ 60$ $\\$ $\Rightarrow A + B = 60 ..... (1)$ $\\$ $tan\ (A + B) = \dfrac{1}{\sqrt{3}}$ $\\$ $\Rightarrow (A - B) = tan\ 30$ $\\$ $\Rightarrow A - B = 30 .... (2)$ $\\$ On adding both equations, we obtain $\\$ $2A = 90$ $\\$ $Rightarrow A = 45$ $\\$ From equation $(1)$, we obtain $\\$ $45 + B = 60$ $\\$ $B = 15$ $\\$ Therefore, $\angle{A} = 45°$ and $\angle{B} = 15°$

15.   State whether the following are true or false. Justify your answer. $\\$ (i) $sin\ (A + B) = sin\ A + sin\ B.$ $\\$ (ii) The value of $sin\ \theta$ increases as $\theta$ increases. $\\$ (iii) The value of $cos\ \theta$ increases as $\theta$ increases. $\\$ (iv) $sin\ \theta = cos\ \theta$ for all values of $\theta$. $\\$ (v) $cot\ A$ is not defined for $A = 0°.$

(i) $sin\ (A + B) = sin\ A + sin\ B$ $\\$ Let $A = 30°$ and $B = 60°$ $\\$ $sin\ (A + B) = sin\ (30° + 60°)$ $\\$ $= sin\ 90°$ $\\$ $= 1$ $\\$ $sin\ A + sin\ B = sin\ 30° + sin\ 60°$ $\\$ $= \dfrac{1}{2} + \dfrac{\sqrt{3}}{2} = \dfrac{1 + \sqrt{3}}{2}$ $\\$ Clearly, $sin\ (A + B) \ne sin\ A + sin\ B$ $\\$ Hence, the given statement is false.

(ii) The value of $sin\ \theta$ increases as $\theta$ increases in the interval of $0° < \theta < 90°$ as $sin\ 0° = 0$ $\\$ $sin\ 30^o = \dfrac{1}{2} = 0.5$ $\\$ $sin\ 45^o = \dfrac{1}{\sqrt{2}} = 0.707$ $\\$ $sin\ 60^o = \dfrac{\sqrt{3}}{2} = 0.866$ $\\$ $sin\ 90° = 1$ $\\$ Hence, the given statement is true.

(iii) $cos\ 0° = 1$ $\\$ $cos\ 90^o = \dfrac{\sqrt{3}}{2} = 0.866$ $\\$ $cos\ 45^o = \dfrac{1}{\sqrt{2}} = 0.707$ $\\$ $cos\ 60^o = \dfrac{1}{2} = 0.5$ $\\$ $cos\ 90° = 0$ $\\$ It can be observed that the value of $cos\ \theta$ does not increase in the interval of $0° < \theta < 90°.$ $\\$ Hence, the given statement is false.

(iv) $sin\ \theta = cos\ \theta$ for all values of $\theta$. This is true when $\theta = 45°$ $\\$ As $sin\ 45^o = \dfrac{1}{\sqrt{2}}$ $\\$ $cos\ 45^o = \dfrac{1}{\sqrt{2}}$ $\\$ It is not true for all other values of $\theta.$ $\\$ As $sin\ 30^o = \dfrac{1}{2}$ and $cos\ 30^o = \dfrac{\sqrt{3}}{2}$, $\\$ Hence, the given statement is false.

(v) $cot\ A$ is not defined for $A = 0°$ $\\$ As $cot\ A = \dfrac{cos\ A}{sin\ A}$ $\\$ $cot\ 0^o = \dfrac{cos\ 0^o}{sin\ 0^o} = \dfrac{1}{0} =$ undefined $\\$ Hence, the given statement is true. $\\$

16.   Evaluate : $\\$ (i) $\dfrac{sin\ 18^ o}{cos\ 72^o}$ $\\$ (ii) $\dfrac{tan\ 26^o}{cot\ 64^o}$ $\\$ (iii) $cos\ 48^o - sin\ 42^o$ $\\$ (iv) $cosec\ 48^o - sin\ 42^o$

(i) $\dfrac{sin\ 18^ o}{cos\ 72^o}$ $\\$ $\dfrac{sin\ 18^ o}{cos\ 72^o} = \dfrac{sin\ (90^o - 72^o)}{cos\ 72^o}$ $\\$ $= \dfrac{cos\ 72^o}{cos\ 72^o} = 1$

(ii) $\dfrac{tan\ 26^o}{cot\ 64^o}$ $\\$ $\dfrac{tan\ 26^o}{cot\ 64^o} = \dfrac{tan\ (90^o - 64^o)}{cot\ 64^o}$ $\\$ $= \dfrac{cot\ 64^o}{cot\ 64^o} = 1$ $\\$

(III) $cos\ 48° - sin\ 42° = cos\ (90°- 42°) - sin\ 42°$ $\\$ $= sin\ 42° - sin\ 42°$ $\\$ $= 0$ $\\$ (IV) $cosec\ 31° - sec\ 59° = cosec\ (90° - 59°) - sec\ 59°$ $\\$ $= sec\ 59° - sec\ 59°$ $\\$ $= 0$

17.   Show that : $\\$ (i) $tan\ 48°\ tan\ 23°\ tan\ 42°\ tan\ 67° = 1$ $\\$ (ii) $cos\ 38°\ cos\ 52° - sin\ 38°\ sin\ 52° = 0$

Answer

17   None

(I) $tan\ 48°\ tan\ 23°\ tan\ 42°\ tan 67°$ $\\$ $= tan\ (90° - 42°)\ tan\ (90° - 67°)\ tan\ 42°\ tan\ 67°$ $\\$ $= cot\ 42°\ cot\ 67°\ tan\ 42°\ tan\ 67°$ $\\$ $= (cot\ 42°\ tan\ 42°) (cot\ 67°\ tan\ 67°)$ $\\$ $= (1) (1)$ $\\$ $= 1$ $\\$ (II) $cos\ 38°\ cos\ 52° - sin\ 38°\ sin\ 52°$ $\\$ $= cos\ (90° - 52°)\ cos\ (90°- 38°) - sin\ 38°\ sin\ 52°$ $\\$ $= sin\ 52°\ sin\ 38° - sin\ 38°\ sin\ 52°$ $\\$ $= 0$

18.   If $tan\ 2A = cot\ (A - 18°)$, where $2A$ is an acute angle, find the value of $A.$

Given that, $\\$ $tan\ 2A = cot\ (A - 18°)$ $\\$ $cot\ (90° - 2A) = cot\ (A - 18°)$ $\\$ $90° - 2A = A - 18°$ $\\$ $108° = 3A$ $\\$ $A = 36°$

19.   If $tan\ A = cot\ B$, prove that $A + B = 90°.$

Given that, $\\$ $tan\ A = cot\ B$ $\\$ $tan\ A = tan\ (90° - B)$ $\\$ $A = 90° - B$ $\\$ $A + B = 90°$

20.   If $sec\ 4A = cosec\ (A - 20°)$, where $4A$ is an acute angle, find the value of $A.$

Given that, $\\$ $sec\ 4A = cosec\ (A - 20°)$ $\\$ $cosec\ (90° - 4A) = cosec\ (A - 20°)$ $\\$ $90° - 4A = A - 20°$ $\\$ $110° = 5A$ $\\$ $A = 22°$

21.   If $A,\ B$ and $C$ are interior angles of a triangle $ABC$, then show that $\\$ $$sin\ \Big(\dfrac{B + C}{2}\Big) = cos\ \dfrac{A}{2}$$

We know that for a triangle $ABC,$ $\\$ $\angle{A} + \angle{B} + \angle{C} = 180°$ $\\$ $\angle{B} + \angle{C} = 180° - \angle{A}$ $\\$ $\dfrac{\angle{B} + \angle{C}}{2} = 90^o - \dfrac{\angle{A}}{2}$ $\\$ $sin \Big(\dfrac{B + C}{2}\Big) = sin \Big(90^o - \dfrac{A}{2}\Big)$ $\\$ $= cos \Big(\dfrac{A}{2}\Big)$ $\\$

22.   Express $sin\ 67° + cos\ 75°$ in terms of trigonometric ratios of angles between $0°$ and $45°.$

$sin\ 67° + cos\ 75°$ $\\$ $= sin\ (90° - 23°) + cos\ (90° - 15°)$ $\\$ $= cos\ 23° + sin\ 15°$ $\\$

23.   Express the trigonometric ratios $sin\ A,\ sec\ A$ and $tan\ A$ in terms of $cot\ A.$

We know that, $\\$ $cosec^2 A = 1 + cot^2 A$ $\\$ $\dfrac{1}{cosec^2 A} = \dfrac{1}{1 + cot^2 A}$ $\\$ $sin^2\ A = \dfrac{1}{1 + cot^2\ A}$ $\\$ $sin\ A = \pm \dfrac{1}{\sqrt{1 + cot^2\ A}}$ $\\$ $\sqrt{1 + cot^2\ A}$ will always be positive as we are adding two positive quantities. $\\$ Therefore, $sin\ A = \dfrac{1}{1 + cot^2\ A}$ $\\$ We know that, $tan\ A = \dfrac{sin\ A}{cos\ A}$ $\\$ However, $cot\ A = \dfrac{cos\ A}{sin\ A}$ $\\$ Therefore, $tan\ A = \dfrac{1}{cot\ A}$ $\\$ Also, $sec^2 A = 1 + tan^2\ A$ $\\$ $= 1 + \dfrac{1}{cot^2\ A}$ $\\$ $= \dfrac{cot^2\ A + 1}{cot^2\ A}$ $\\$

24.   Write all the other trigonometric ratios of $\angle{A}$ in terms of $sec\ A.$

We know that, $\\$ $cos\ A = \dfrac{1}{sec\ A}$ $\\$ Also $sin^2\ A + cos^2\ A = 1$ $\\$ $sin^2\ A = 1 - cos^2\ A$ $\\$ $sin\ A = \sqrt{1 - (\dfrac{1}{sec\ A})^2}$ $\\$ $= \sqrt{\dfrac{sec^2\ A - 1}{sec^2\ A}} = \dfrac{\sqrt{sec^2\ A - 1}}{sec\ A}$ $\\$ $tan^2A + 1 = sec^2A$ $\\$ $tan^2A = sec^2A - 1$ $\\$ $tan\ A = \sqrt{sec^2\ A - 1}$ $\\$ $cot\ A = \dfrac{cos\ A}{sin\ A} = \dfrac{\dfrac{1}{sec\ A}}{\dfrac{\sqrt{sec^2\ A - 1}}{sec\ A}}$ $\\$ $= \dfrac{1}{\sqrt{sec^2\ A - 1}}$ $\\$ $cosec\ A = \dfrac{1}{sin\ A} = \dfrac{sec\ A}{\sqrt{sec^2\ A - 1}}$ $\\$

25.   Evaluate : $\\$ (i) $\dfrac{sin^2\ 63^o + sin^2\ 27^o}{cos^2\ 17^o + cos^2\ 73^o}$ $\\$ (ii) $sin\ 25°\ cos\ 65° + cos\ 25°\ sin\ 65°$

(i) $\dfrac{sin^2\ 63^o + sin^2\ 27^o}{cos^2\ 17^o + cos^2\ 73^o}$ $\\$ $= \dfrac{[sin\ (90^o - 27^o)] + sin^2\ 27^o}{[sin\ (90^o - 27^o)] + cos^2\ 73^o}$ $\\$ $= \dfrac{[cos\ 27^o] + sin^2\ 27^o}{[sin^2\ 73^o]^2 + cos^2\ 73^o}$ $\\$ $ = \dfrac{cos^ 27^o + sin^2\ 27^o}{sin^2\ 73^o + cos^2 73^o}$ $\\$ $= \dfrac{1}{1}(As\ sin^2 A + Cos^2\ A = 1)$ $\\$ $= 1$ $\\$

(ii) $sin\ 25°\ cos\ 65° + cos\ 25°\ sin\ 65°$ $\\$ $= (sin\ 25){cos(90^o - 25^o)} + cos\ 25^o {sin(90^o - 25^o)}$ $\\$ $= (sin\ 25^o)(sin\ 25^o) + (cos\ 25^o)(cos\ 25^o)$ $\\$ $= sin^225° + cos^225°$ $= 1 (As\ sin2A + cos2A = 1)$

26.   Evaluate : $\\$ (i) $\dfrac{sin^2\ 63^o + sin^2\ 27^o}{cos^2\ 17^o + cos^2\ 73^o}$ $\\$ (ii) $sin\ 25°\ cos\ 65° + cos\ 25°\ sin\ 65°$

27.   Choose the correct option. Justify your choice. $\\$ (i) $9\ sec^2\ A - 9\ tan^2\ A =$ $\\$ $(A)\ 1 \ \ \ \ \ \ \ \ (B)\ 9 \ \ \ \ \ \ \ \ (C)\ 8 \ \ \ \ \ \ \ \ (D)\ 0$ $\\$ (ii) $(1 + tan\ \theta + sec\ \theta)(1 + cot\ \theta - cosec\ \theta) =$ $\\$ $(A)\ 0 \ \ \ \ \ \ \ \ (B)\ 1 \ \ \ \ \ \ \ \ (C)\ 2 \ \ \ \ \ \ \ \ (D)\ -1$ $\\$ (iii) $(sec\ A + tan\ A) (1 - sin\ A) =$ $\\$ $(A)\ sec\ A \ \ \ \ \ \ \ \ (B)\ sin\ A \ \ \ \ \ \ \ \ (C)\ cosec\ A \ \ \ \ \ \ \ \ (D)\ cos\ A$ $\\$ (iv) $\dfrac{1 + tan^2\ A}{1 + cot^2\ A} =$ $\\$ $(A)\ sec^2\ A \ \ \ \ \ \ \ \ (B)\ -1 \ \ \ \ \ \ \ \ (C)\ cot^2\ A \ \ \ \ \ \ \ \ (D)\ tan^2\ A$

Answer

27   None

(i) $9 sec^2A - 9\ tan^2A$ $\\$ $= 9\ (sec^2A - tan^2A)$ $\\$ $= 9\ (1)\ [As\ sec^2 A - tan^2\ A = 1]$ $\\$ $= 9$ $\\$ Hence, alternative $(B)$ is correct. $\\$

(ii) $(1 + tan \theta + sec\ \theta)\ (1 + cot\ \theta - cosec\ \theta)$ $\\$ $= \Big(1 + \dfrac{sin\ \theta}{cos\ \theta} + \dfrac{1}{cos\ \theta}\Big) \Big(1 + \dfrac{cos\ \theta}{sin\ \theta} - \dfrac{1}{sin\ \theta}\Big)$ $\\$ $= \Big(\dfrac{cos\ \theta + sin\ \theta + 1}{cos\ \theta}\Big) \Big(\dfrac{sin\ \theta + cos\ \theta - 1}{sin\ \theta}\Big)$ $\\$ $= \dfrac{(sin\ \theta + sin\ \theta)^2 - (1)^2}{sin\ \theta\ cos\ \theta}$ $\\$ $= \dfrac{sin^2\ \theta + cos^2\ \theta + 2\ sin\ \theta\ cos\ \theta - 1}{sin\ \theta\ cos\ \theta}$ $\\$ $= \dfrac{1 + 2\ sin\ \theta\ cos\ \theta - 1}{sin\ \theta\ cos\ \theta}$ $\\$ $= \dfrac{2\ sin\ \theta\ cos\ \theta}{sin\ \theta\ cos\ \theta} = 2$ $\\$ Hence, alternative $(C)$ is correct.

(iii) $(sec\ A + tan\ A) (1 - sin\ A) =$ $\\$ $= \Big(\dfrac{1}{cos\ A} + \dfrac{sin\ A}{cos\ A}\Big)(1 - sin\ A)$ $\\$ $= \Big(\dfrac{1 + sin\ A}{cos\ A}\Big)(1 - sin\ A)$ $\\$ $= \dfrac{1 - sin^2\ A}{cos\ A} = \dfrac{cos^2\ A}{cos\ A}$ $\\$ $= cos\ A$ $\\$ Hence, alternative $(D)$ is correct.

(iv) $\dfrac{1 + tan^2\ A}{1 + cot^2\ A} =$ $\\$ $\dfrac{ 1 + \dfrac{sin^2\ A}{cos^2\ A}}{1 + \dfrac{cos^2\ A}{sin^2\ A}}$ $\\$ $=\dfrac{\dfrac{cos^2\ A + sin^2\ A}{cos^2\ A}}{\dfrac{sin^2\ A + cos^2\ A}{sin^2\ A}} = \dfrac{\dfrac{1}{cos^2\ A}}{\dfrac{1}{sin^2\ A}}$ $\\$ $= \dfrac{sin^2\ A}{cos^2\ A} = tan^2\ A$ $\\$ Hence, alternative $(D)$ is correct.

28.   Prove the following identities, where the angles involved are acute angles for which the expressions are defined. $\\$ (i) $(cosec\ \theta - cot\ \theta)^2 = \dfrac{1 - cos\ \theta}{1 + cos\ \theta}$ $\\$ (ii) $\dfrac{cos\ A}{1 + sin\ A} + \dfrac{1 + sin\ A}{cos\ A} = 2\ sec\ A$ $\\$ (iii) $\dfrac{tan\ \theta}{1 - cot\ \theta} + \dfrac{cot\ \theta}{1 - tan\ \theta} = 1 + sec\ \theta\ cosec\ \theta$ $\\$ [$Hint :$ Write the expression in terms of $sin\ \theta$ and $cos\ \theta$] $\\$ (iv) $\dfrac{1 + sec\ A}{sec\ A} = \dfrac{sin^2\ A}{1 - cos\ A} \ \ \ \ \ \ $ [$Hint :$ Simplify $LHS$ and $RHS$ separately] $\\$ (v) $\dfrac{cos\ A - sin\ A + 1}{cos\ A + sin\ A - 1} = cosec\ A + cot\ A,$ using the identity $cosec^2\ A = 1 + cot^2\ A.$ $\\$ (vi) $\sqrt{\dfrac{1 + sin\ A}{1 - sin\ A}} = sec\ A + tan\ A$ $\\$ (vii) $ \dfrac{sin\ \theta - 2\ sin^3\ \theta}{2\ cos^3 \theta - cos\ \theta} = tan\ \theta$ $\\$ (viii) $(sin\ A + cosec\ A)^2 + (cos\ A + sec\ A)^2 = 7 + tan^2\ A + cot^2\ A$ $\\$ (ix) $(cosec\ A - sin\ A) (sec\ A - cos\ A) = \dfrac{1}{tan\ A + cot\ A}$ $\\$ $ \ \ \ \ \ \ \ $ [$Hint :$ Simplify $LHS$ and $RHS$ separately] $\\$ (x) $\Big(\dfrac{1 + tan^2\ A}{1 + cot^2\ A}\Big) = \Big(\dfrac{1 - tan\ A}{1 - cot\ A}\Big)^2 = tan^2\ A$

(i) $(cosec\ \theta - cot\ \theta)^2 = \dfrac{1 - cos\ \theta}{1 + cos\ \theta}$ $\\$ $L.H.S = (cosec\ \theta - cot\ \theta)^2$ $\\$ $= \Big(\dfrac{1}{sin\ \theta} - \dfrac{cos\ \ theta}{sin\ \theta}\Big)$ $\\$ $= \dfrac{(1 - cos\ \theta)^2}{(sin\ \theta)^2} = \dfrac{(1 - cos\ \theta)^2}{sin^2\ \theta}$ $\\$ $= \dfrac{(1 - cos\ \theta)^2}{(1 - cos^2\ \theta)} = \dfrac{(1 - cos\ \theta)^2}{(1 - cos\ \theta)(1 + cos\ \theta)} = \dfrac{1 - cos\ \theta}{1 + cos\ \theta}$ $\\$ $= R.H.S$

(ii) $\dfrac{cos\ A}{1 + sin\ A} + \dfrac{1 + sin\ A}{cos\ A} = 2\ sec\ A$ $\\$ $L.H.S = \dfrac{cos\ A}{1 + sin\ A} + \dfrac{1 + sin\ A}{cos\ A}$ $\\$ $= \dfrac{cos^2\ A + (1 + sin\ A)^2}{(1 + sin\ A)(cos\ A)}$ $\\$ $\dfrac{cos^2\ A + 1 + sin^2\ A + 2\ sin\ A}{(1 + sin\ A)(cos\ A)}$ $\\$ $= \dfrac{sin^2\ A + cos^2\ A + 1 + 2\ sin\ A}{(1 + sin\ A)(cos\ A)}$ $\\$ $= \dfrac{1 + 1 + 2\ sin\ A}{(1 + sin\ A)(cos\ A)} = \dfrac{2 + 2\ sin\ A}{(1 + sin\ A)(cos\ A)}$ $\\$ $= \dfrac{2(1 + sin\ A)}{(1 + sin\ A)(cos\ A)} = \dfrac{2}{cos\ A} =2\ sec\ A$ $\\$ $= R.H.S$

(iii) $\dfrac{tan\ \theta}{1 - cot\ \theta} + \dfrac{cot\ \theta}{1 - tan\ \theta} = 1 + sec\ \theta\ cosec\ \theta$ $\\$ $L.H.S = \dfrac{tan\ \theta}{1 - cot\ \theta} + \dfrac{cot\ \theta}{1 - tan\ \theta}$ $\\$ $= \dfrac{\dfrac{sin\ \theta}{cos\ \theta}}{1 - \dfrac{cos\ \ theta}{sin\ \theta}} + \dfrac{\dfrac{cos\ \theta}{sin\ \theta}}{1 - \dfrac{sin\ \theta}{cos\ \theta}}$ $\\$ $= \dfrac{\dfrac{sin\ \theta}{cos\ \theta}}{\dfrac{sin\ \theta - cos\ \theta}{sin\ \theta}} + \dfrac{\dfrac{cos\ \theta}{sin\ \theta}}{\dfrac{cos\ \theta - sin\ \theta}{cos\ \theta}}$ $\\$ $= \dfrac{sin^2\ \theta}{cos\ \theta(sin\ \theta - cos\ \theta)} + \dfrac{cos^2\ \theta}{sin\ \theta(sin\ \theta - cos\ \theta)}$ $\\$ $= \dfrac{1}{(sin\ \theta - cos\ \theta)} \Big[\dfrac{sin^2\ \theta}{cos\ \theta} - \dfrac{cos^2\ \theta}{sin\ \theta}\Big]$ $\\$ $= \Big(\dfrac{1}{sin\ \theta}\Big)$