# Class 10 NCERT Introduction to Trigonometry

#### 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$$\\$

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$(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)\$