Class 10: Trigonometry – Exercise 23(c) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note: Trigonometrical Ratios of complementary angles:

For $\displaystyle \theta < 90^{\circ}$

(i) $\displaystyle \sin (90^{\circ} - \theta) = \cos \theta$ (ii) $\displaystyle \cos (90^{\circ} - \theta) = \sin \theta$

(iii) $\displaystyle \tan (90^{\circ} - \theta) = \cot \theta$ (iv) $\displaystyle \cot (90^{\circ} - \theta) = \tan \theta$

(v) $\displaystyle \sec (90^{\circ} - \theta) = \mathrm{cosec} \theta$ (vi) $\displaystyle \mathrm{cosec} (90^{\circ} - \theta) = \sec \theta$

Evaluate:

$\displaystyle \text{Question 1: } \frac{\cos 22^{\circ}}{\sin 68^{\circ}}$

Answer:

$\displaystyle \frac{\cos 22^{\circ}}{\sin 68^{\circ}} = \frac{\cos (90^{\circ}-68^{\circ})}{\sin 68^{\circ}} = \frac{\sin 68^{\circ}}{\sin 68^{\circ}} = 1$

$\displaystyle \\$

$\displaystyle \text{Question 2: } \frac{\tan 47^{\circ}}{\cot 43^{\circ}}$

Answer:

$\displaystyle \frac{\tan 47^{\circ}}{\cot 43^{\circ}} = \frac{\tan (90^{\circ}-43^{\circ})}{\cot 43^{\circ}} = \frac{\cot 43^{\circ}}{\cot 43^{\circ}} = 1$

$\displaystyle \\$

$\displaystyle \text{Question 3: } \frac{\sec 75^{\circ}}{\mathrm{cosec} 15^{\circ}}$

Answer:

$\displaystyle \frac{\sec 75^{\circ}}{\mathrm{cosec} 15^{\circ}} = \frac{\sec (90^{\circ}-15^{\circ})}{\mathrm{cosec} 5^{\circ}} = \frac{\mathrm{cosec} 15^{\circ}}{\mathrm{cosec} 15^{\circ}} = 1$

$\displaystyle \\$

$\displaystyle \text{Question 4: } \frac{\cos 55^{\circ}}{\sin 35^{\circ}}+\frac{\cot 35^{\circ}}{\tan 55^{\circ}}$

Answer:

$\displaystyle \frac{\cos 55^{\circ}}{\sin 35^{\circ}}+\frac{\cot 35^{\circ}}{\tan 55^{\circ}} = \frac{\cos (90-35)}{\sin 35^{\circ}}+\frac{\cot (90-35)^{\circ}}{\tan 55^{\circ}} = \frac{\sin 35^{\circ}}{\sin 35^{\circ}}+\frac{\tan 55^{\circ}}{\tan 55^{\circ}} = 2$

$\displaystyle \\$

$\displaystyle \text{Question 5: } \cos^2 40^{\circ} + \cos^2 50^{\circ}$

Answer:

$\displaystyle \cos^2 40^{\circ} + \cos^2 50^{\circ} = (\cos (90^{\circ}-50^{\circ}))^2 + \cos^2 50^{\circ} = \sin^2 50^{\circ} + \cos^2 50^{\circ} = 1$

$\displaystyle \\$

$\displaystyle \text{Question 6: } \sec^2 18^{\circ}- \cot^2 72^{\circ}$

Answer:

$\displaystyle \sec^2 18^{\circ}- \cot^2 72^{\circ} = (\sec (90^{\circ}-72^{\circ}))^2 - \cot^2 72^{\circ} = \mathrm{cosec}^2 72^{\circ} - \cot^2 72^{\circ} = 1$

$\displaystyle \\$

$\displaystyle \text{Question 7: } \sin 15^{\circ} . \cos 75^{\circ} + \cos 15^{\circ}.\sin 75^{\circ}$

Answer:

$\displaystyle \sin 15^{\circ} . \cos 75^{\circ} + \cos 15^{\circ}.\sin 75^{\circ}$

$\displaystyle = \sin (90^{\circ} - 75^{\circ}) . \cos 75^{\circ} + \cos (90^{\circ} - 75^{\circ}).\sin 75^{\circ}$

$\displaystyle = \cos^2 75 + \sin^2 75 = 1$

$\displaystyle \\$

$\displaystyle \text{Question 8: } \sin 42^{\circ} . \sin 48^{\circ} - \cos 42^{\circ}.\cos 48^{\circ}$

Answer:

$\displaystyle \sin 42^{\circ} . \sin 48^{\circ} - \cos 42^{\circ}.\cos 48^{\circ}$

$\displaystyle = \sin 42^{\circ} . \sin (90^{\circ} - 42^{\circ}) - \cos 42^{\circ}.\cos (90^{\circ} - 42^{\circ})$

$\displaystyle = \sin 42^{\circ}.\cos 42^{\circ} - \cos 42^{\circ}. \sin 42^{\circ} = 0$

$\displaystyle \\$

$\displaystyle \text{Question 9: } \sin (90^{\circ}-A) \cos A + \cos (90^{\circ}-A) \sin A$

Answer:

$\displaystyle \sin (90^{\circ}-A) \cos A + \cos (90^{\circ}-A) \sin A = \cos^2 A + \sin^2 A = 1$

$\displaystyle \\$

$\displaystyle \text{Question 10: } \displaystyle \sin^2 35^{\circ} + \sin^2 55^{\circ}$

Answer:

$\displaystyle \sin^2 35^{\circ} + \sin^2 55^{\circ} = \{\sin (90-55)\}^2 + \sin^2 55^{\circ} = \cos^2 55^{\circ} + \sin ^2 55^{\circ} = 1$

$\displaystyle \\$

$\displaystyle \text{Question 11: }\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}} - 2$

Answer:

$\displaystyle \frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}$ $\displaystyle = \frac{\cot (90^{\circ} - 36^{\circ})}{\tan 36^{\circ}}+\frac{\tan (90^{\circ} - 70^{\circ})}{\cot 70^{\circ}}$ $\displaystyle = \frac{\tan 36^{\circ}}{\tan 36^{\circ}}+\frac{\cot 70^{\circ}}{\cot 70^{\circ}} = 2$

$\displaystyle \\$

$\displaystyle \text{Question 12: }\frac{2 \tan 53^{\circ}}{\cot 37^{\circ}}-\frac{\cot 80^{\circ}}{\tan 10^{\circ}} \hspace{1.0cm} [2006]$

Answer:

$\displaystyle \frac{2 \tan 53^{\circ}}{\cot 37^{\circ}}-\frac{\cot 80^{\circ}}{\tan 10^{\circ}}$ $\displaystyle = \frac{2 \tan (90^{\circ} - 37^{\circ})}{\cot 37^{\circ}}-\frac{\cot (90^{\circ} - 10^{\circ})}{\tan 10^{\circ}}$ $\displaystyle = \frac{2 \cot 37^{\circ}}{\cot 37^{\circ}}-\frac{\tan 10^{\circ}}{\tan 10^{\circ}} = 0$

$\displaystyle \\$

$\displaystyle \text{Question 13: }\cos^2 25^{\circ} + \cos^2 65^{\circ} - \tan^2 45^{\circ}$

Answer:

$\displaystyle \cos^2 25^{\circ} + \cos^2 65^{\circ} - \tan^2 45^{\circ}$

$\displaystyle = \cos^2 (90^{\circ}-65^{\circ}) + \cos^2 65^{\circ} - \tan^2 45^{\circ}$

$\displaystyle = \sin^2 65^{\circ} + \cos^2 65^{\circ} - \tan^2 45^{\circ} = 0$

$\displaystyle \\$

$\displaystyle \text{Question 14: }\frac{\cos^2 32^{\circ} + \cos^2 58^{\circ}}{\sin^2 59^{\circ} + \sin^2 31^{\circ}}$

Answer:

$\displaystyle \frac{\cos^2 32^{\circ} + \cos^2 58^{\circ}}{\sin^2 59^{\circ} + \sin^2 31^{\circ}}$ $\displaystyle = \frac{\cos^2 (90^{\circ} - 58^{\circ}) + \cos^2 58^{\circ}}{\sin^2 (90^{\circ} - 31^{\circ}) + \sin^2 31^{\circ}}$ $\displaystyle = \frac{\sin^2 58^{\circ} + \cos^2 58^{\circ}}{\cos^2 31^{\circ} + \sin^2 31^{\circ}} = 1$

$\displaystyle \\$

$\displaystyle \text{Question 15: }(\frac{\sin 77^{\circ}}{\cos 13^{\circ}})^2 + (\frac{\cos 77^{\circ}}{\sin 13^{\circ}})^2 - 2 \cos^2 45^{\circ}$

Answer:

$\displaystyle (\frac{\sin 77^{\circ}}{\cos 13^{\circ}})^2 + (\frac{\cos 77^{\circ}}{\sin 13^{\circ}})^2 - 2 \cos^2 45^{\circ}$

$\displaystyle = (\frac{\sin (90^{\circ} - 13^{\circ})}{\cos 13^{\circ}})^2 + (\frac{\cos (90^{\circ} - 13^{\circ})}{\sin 13^{\circ}})^2 - 2 \cos^2 45^{\circ}$

$\displaystyle = (\frac{\cos 13^{\circ}}{\cos 13^{\circ}})^2 + (\frac{\sin 13^{\circ}}{\sin 13^{\circ}})^2 - 2 \cos^2 45^{\circ}$

$\displaystyle = 1 + 1 - 2 \times (\frac{1}{\sqrt{2}})^2 = 1$

$\displaystyle \\$

$\displaystyle \text{Question 16: }\cos^2 26^{\circ} + \cos 64^{\circ}.\sin 26^{\circ} + \frac{\tan 36^{\circ}}{\cot 54^{\circ}} \hspace{1.0cm} [2012]$

Answer:

$\displaystyle \cos^2 26^{\circ} + \cos 64^{\circ}.\sin 26^{\circ} + \frac{\tan 36^{\circ}}{\cot 54^{\circ}}$

$\displaystyle = \cos^2 26^{\circ} + \cos (90^{\circ} - 26^{\circ}).\sin 26^{\circ} + \frac{\tan (90^{\circ} - 54^{\circ})}{\cot 54^{\circ}}$

$\displaystyle = \cos^2 26^{\circ} + \sin^2 26^{\circ} + \frac{\cot 54^{\circ}}{\cot 54^{\circ}}$

$\displaystyle = 2$

$\displaystyle \\$

$\displaystyle \text{Question 17: } 3 .\frac{\sin 72^{\circ}}{\cos 18^{\circ}} - \frac{\sec 32^{\circ}}{\mathrm{cosec} 58^{\circ}}$

Answer:

$\displaystyle 3 .\frac{\sin 72^{\circ}}{\cos 18^{\circ}} - \frac{\sec 32^{\circ}}{\mathrm{cosec} 58^{\circ}}$

$\displaystyle = 3 .\frac{\sin (90^{\circ} - 18^{\circ})}{\cos 18^{\circ}} - \frac{\sec (90^{\circ} - 32^{\circ})}{\mathrm{cosec} 58^{\circ}}$

$\displaystyle = 3 .\frac{\cos 18^{\circ}}{\cos 18^{\circ}} - \frac{\mathrm{cosec} 58^{\circ}}{\mathrm{cosec} 58^{\circ}} = 3 - 1 = 2$

$\displaystyle \\$

$\displaystyle \text{Question 18: } 3 \cos 80^{\circ} . \mathrm{cosec} 10^{\circ} + 2 \sin 59^{\circ}.\sec 31^{\circ} \hspace{1.0cm} [2013]$

Answer:

$\displaystyle 3 \cos 80^{\circ} . \mathrm{cosec} 10^{\circ} + 2 \sin 59^{\circ}.\sec 31^{\circ}$

$\displaystyle = 3 \cos 80^{\circ} . \mathrm{cosec} (90^{\circ} - 80^{\circ}) + 2 \sin 59^{\circ}.\sec (90^{\circ} - 59^{\circ})$

$\displaystyle = 3 \cos 80^{\circ} . \sec 80^{\circ} + 2 \sin 59^{\circ} . \mathrm{cosec} 59^{\circ}$

$\displaystyle = 3 + 2 = 5$

$\displaystyle \\$

$\displaystyle \text{Question 19: } \frac{\sin 80^{\circ}}{\cos 10^{\circ}} + \sin 59^{\circ} . \sec 31^{\circ} \hspace{1.0cm} [2007]$

Answer:

$\displaystyle \frac{\sin 80^{\circ}}{\cos 10^{\circ}} + \sin 59^{\circ} . \sec 31^{\circ}$

$\displaystyle = \frac{\sin (90^{\circ} - 10^{\circ})}{\cos 10^{\circ}} + \sin 59^{\circ} . \sec (90^{\circ} - 59^{\circ})$

$\displaystyle = \frac{\cos 10^{\circ}}{\cos 10^{\circ}} + \sin 59^{\circ} . \mathrm{cosec} 59^{\circ} = 2$

$\displaystyle \\$

$\displaystyle \text{Question 20: } \tan (55^{\circ}-A) - \cot (35^{\circ}+A)$

Answer:

$\displaystyle \tan (55^{\circ}-A) - \cot (35^{\circ}+A)$

$\displaystyle = \tan ((90^{\circ} -(35^{\circ}+A)) - \cot (35^{\circ}+A)$

$\displaystyle = \cot (35^{\circ}+A) - \cot (35^{\circ}+A) = 0$

$\displaystyle \\$

$\displaystyle \text{Question 21: } \mathrm{cosec} (65^{\circ}+A) - \sec (25^{\circ}-A)$

Answer:

$\displaystyle \mathrm{cosec} (65^{\circ}+A) - \sec (25^{\circ}-A)$

$\displaystyle = \mathrm{cosec} (90^{\circ} - (25^{\circ}-A)) - \sec (25^{\circ}-A)$

$\displaystyle = \sec (25^{\circ}-A) - \sec (25^{\circ}-A) = 0$

$\displaystyle \\$

$\displaystyle \text{Question 22: } 2 .\frac{\tan 57^{\circ}}{\cot 33^{\circ}} - \frac{\cot 70^{\circ}}{\tan 20^{\circ}} - \sqrt{2} \cos 45^{\circ}$

Answer:

$\displaystyle 2 .\frac{\tan 57^{\circ}}{\cot 33^{\circ}} - \frac{\cot 70^{\circ}}{\tan 20^{\circ}} - \sqrt{2} \cos 45^{\circ}$

$\displaystyle = 2 .\frac{\tan (90^{\circ} - 33^{\circ})}{\cot 33^{\circ}} - \frac{\cot (90^{\circ} - 20^{\circ})}{\tan 20^{\circ}} - \sqrt{2} \cos 45^{\circ}$

$\displaystyle = 2 .\frac{\cot 33^{\circ}}{\cot 33^{\circ}} - \frac{\tan 20^{\circ}}{\tan 20^{\circ}} - \sqrt{2} \cos 45^{\circ}$

$\displaystyle = 2 - 1 - \sqrt{2} \times \frac{1}{\sqrt{2}}$

$\displaystyle = 0$

$\displaystyle \\$

$\displaystyle \text{Question 23: } \frac{\cot^2 41^{\circ}}{\tan^2 49^{\circ}} - 2 \frac{\sin^2 75^{\circ}}{\cos^2 15^{\circ}}$

Answer:

$\displaystyle \frac{\cot^2 41^{\circ}}{\tan^2 49^{\circ}} - 2 \frac{\sin^2 75^{\circ}}{\cos^2 15^{\circ}}$

$\displaystyle = \frac{\cot^2 (90^{\circ} - 49^{\circ})}{\tan^2 49^{\circ}} - 2 \frac{\sin^2 (90^{\circ} - 15^{\circ})}{\cos^2 15^{\circ}}$

$\displaystyle = \frac{\tan^2 49^{\circ}}{\tan^2 49^{\circ}} - 2 \frac{\cos^2 15^{\circ}}{\cos^2 15^{\circ}} =-1$

$\displaystyle \\$

$\displaystyle \text{Question 24: } \frac{\cos 70^{\circ}}{\sin 20^{\circ}} + \frac{\cos 59^{\circ}}{\sin 31^{\circ}} - 8 \sin^2 30^{\circ}$

Answer:

$\displaystyle \frac{\cos 70^{\circ}}{\sin 20^{\circ}} + \frac{\cos 59^{\circ}}{\sin 31^{\circ}} - 8 \sin^2 30^{\circ}$

$\displaystyle = \frac{\cos (90^{\circ} - 20^{\circ})}{\sin 20^{\circ}} + \frac{\cos (90^{\circ} - 31^{\circ})}{\sin 31^{\circ}} - 8 \sin^2 30^{\circ}$

$\displaystyle = \frac{\sin 20^{\circ}}{\sin 20^{\circ}} + \frac{\sin 31^{\circ}}{\sin 31^{\circ}} - 8 \times \frac{1}{4}$

$\displaystyle = 0$

$\displaystyle \\$

$\displaystyle \text{Question 25: } 14 \sin 30^{\circ} + 6 \cos 60^{\circ} - 5 \tan 45^{\circ} \hspace{1.0cm} [2004]$

Answer:

$\displaystyle 14 \sin 30^{\circ} + 6 \cos 60^{\circ} - 5 \tan 45^{\circ}$

$\displaystyle = 14 \sin (90^{\circ} - 60^{\circ}) + 6 \cos 60^{\circ} - 5 \tan 45^{\circ}$

$\displaystyle = 14 \cos 60^{\circ} + 6 \cos 60^{\circ} - 5 \tan 45^{\circ}$

$\displaystyle = 20 \cos 60^{\circ} - 5 \tan 45^{\circ}$

$\displaystyle = 20 \times \frac{1}{2} - 5 \times 1$

$\displaystyle = 10-5 = 5$

$\displaystyle \\$

$\displaystyle \text{Question 26: Show that: } \tan 10^{\circ} \tan 15^{\circ} \tan 75^{\circ} \tan 80^{\circ} = 1$

Answer:

$\displaystyle \tan 10^{\circ} \tan 15^{\circ} \tan 75^{\circ} \tan 80^{\circ}$

$\displaystyle = \tan (90^{\circ} - 80^{\circ}) \tan (90^{\circ} - 75^{\circ}) \tan 75^{\circ} \tan 80^{\circ}$

$\displaystyle = \cot 80^{\circ} \cot 75^{\circ} \tan 75^{\circ} \tan 80^{\circ}$

$\displaystyle = 1$ . Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 27: Show that: } \sin 42^{\circ} .\sec 48^{\circ} + \cos 42^{\circ} .\mathrm{cosec} 48^{\circ} = 2$

Answer:

$\displaystyle \sin 42^{\circ} .\sec 48^{\circ} + \cos 42^{\circ} .\mathrm{cosec} 48^{\circ}$

$\displaystyle = \sin 42^{\circ} .\sec (90^{\circ} - 42^{\circ}) + \cos 42^{\circ} .\mathrm{cosec} (90^{\circ} - 42^{\circ})$

$\displaystyle = \sin 42^{\circ} .\mathrm{cosec} 42^{\circ} + \cos 42^{\circ} .\sec 42^{\circ}$

$\displaystyle = 1 + 1 = 2.$ Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 28: Show that: } \frac{\sin 26^{\circ}}{\sec 64^{\circ}} + \frac{\cos 26^{\circ}}{\mathrm{cosec} 64^{\circ}} = 2$

Answer:

$\displaystyle \frac{\sin 26^{\circ}}{\sec 64^{\circ}} + \frac{\cos 26^{\circ}}{\mathrm{cosec} 64^{\circ}}$

$\displaystyle = \frac{\sin 26^{\circ}}{\sec (90^{\circ} - 26^{\circ})} + \frac{\cos 26^{\circ}}{\mathrm{cosec} (90^{\circ} - 26^{\circ})}$

$\displaystyle = \frac{\sin 26^{\circ}}{\mathrm{cosec} 26^{\circ}} + \frac{\cos 26^{\circ}}{\sec 26^{\circ}}$

$\displaystyle = \sin^2 26^{\circ} + \cos^2 26^{\circ} = 1 .$ . Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 29: Express in terms of angles between } 0^{\circ}$ and $\displaystyle 45^{\circ} \text{ : } \sin 59^{\circ} + \tan 63^{\circ}$

Answer:

$\displaystyle \sin 59^{\circ} + \tan 63^{\circ} = \sin (90^{\circ} - 31^{\circ}) + \tan (90^{\circ} - 27^{\circ}) = \cos 31^{\circ} + \cot 27^{\circ}$

$\displaystyle \\$

$\displaystyle \text{Question 30: Express in terms of angles between } 0^{\circ}$ and $\displaystyle 45^{\circ} \text{ : } \mathrm{cosec} 68^{\circ} + \cot 72^{\circ}$

Answer:

$\displaystyle \mathrm{cosec} 68^{\circ} + \cot 72^{\circ} = \mathrm{cosec} (90^{\circ} - 32^{\circ}) + \cot (90^0 - 28^{\circ}) = \sec 32^{\circ} + \tan 28^{\circ}$

$\displaystyle \\$

$\displaystyle \text{Question 31: Express in terms of angles between } 0^{\circ}$ and $\displaystyle 45^{\circ} \text{ : } \cos 74^{\circ} + \sec 67^{\circ}$

Answer:

$\displaystyle \cos 74^{\circ} + \sec 67^{\circ} = \cos (90^{\circ} - 16^{\circ}) + \sec (90^{\circ} - 23^{\circ}) = \sin 16^{\circ} + \mathrm{cosec} 23^{\circ}$

$\displaystyle \\$

$\displaystyle \text{Question 32: } \frac{\sin A}{\sin(90^{\circ}-A)} + \frac{\cos A}{\cos(90^{\circ}-A)} = \sec A \mathrm{cosec} A$

Answer:

$\displaystyle \text{LHS } = \frac{\sin A}{\sin(90^{\circ}-A)} + \frac{\cos A}{\cos(90^{\circ}-A)}$

$\displaystyle = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}$

$\displaystyle = \frac{\sin^2 A + \cos^2 A}{\sin A. \cos A}$

$\displaystyle = \sec A \mathrm{cosec} A$ = RHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{Question 33: } \sin A \cos A - \frac{\sin A \cos(90^{\circ}-A) \cos A}{\sec(90^{\circ} - A)} - \frac{\cos A \sin(90^{\circ}-A) \sin A}{\mathrm{cosec} (90^{\circ} - A)} = 0$

Answer:

$\displaystyle \text{LHS } = \sin A \cos A - \frac{\sin A \cos(90^{\circ}-A) \cos A}{\sec(90^{\circ} - A)} - \frac{\cos A \sin(90^{\circ}-A) \sin A}{\mathrm{cosec} (90^{\circ} - A)}$

$\displaystyle = \sin A \cos A - \frac{\sin^2 A \cos A}{\mathrm{cosec} A} - \frac{\cos^2 A \sin A}{\sec A}$

$\displaystyle = \sin A \cos A - \sin^3 A \cos A - \cos^3 A \sin A$

$\displaystyle = \sin A \cos A - \sin A \cos A(\sin^2 A + \cos^2 A)$

$\displaystyle = \sin A \cos A - \sin A \cos A = 0 = \text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 34: For } \triangle ABC \text{, show that: } \sin (\frac{A + B}{2}) = \cos (\frac{C}{2})$

Answer:

$\displaystyle \text{LHS } = \sin (\frac{A + B}{2}) = \sin (\frac{180 - C}{2}) = \sin (90 - \frac{C}{2}) = \cos \frac{C}{2} = \text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 35: For } \triangle ABC \text{, show that: } \tan (\frac{B + C}{2}) = \cot (\frac{A}{2})$

Answer:

$\displaystyle \text{LHS } = \tan (\frac{B+C}{2}) = \tan (\frac{180 - A}{2}) = \tan (90 - \frac{A}{2}) = \cot \frac{A}{2} = \text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \triangle \text{Question 36: In } ABC \text{ is the right angles at B. Find the value of: } \\ \\ \frac{\sec A . \mathrm{cosec} A - \tan A . \cot C}{\sin B}$

Answer:

$\displaystyle \frac{\sec A . \mathrm{cosec} A - \tan A . \cot C}{\sin B}$

$\displaystyle = \frac{\sec A . \mathrm{cosec} (90^0 -A) - \tan A . \cot (90^0 -A)}{\sin 90^{\circ}}$

$\displaystyle = \frac{\sec A . \sec A - \tan A . \tan A}{1}$

$\displaystyle = \sec^2 A - \tan^2 A$

$\displaystyle = \frac{1 - \sin^2 A}{\cos^2 A}$

$\displaystyle = 1$

$\displaystyle \\$

Question 37: Find $\displaystyle x$ if: $\displaystyle \sin x = \sin 60^{\circ} \cos 30^{\circ} - \cos 60^{\circ} \sin 30^{\circ}$

Answer:

$\displaystyle \sin x = \sin 60^{\circ} \cos 30^{\circ} - \cos 60^{\circ} \sin 30^{\circ}$

$\displaystyle \Rightarrow \sin x = \sin 60^{\circ} \cos (90^{\circ} - 60^{\circ}) - \cos 60^{\circ} \sin (90^{\circ} - 60^{\circ})$

$\displaystyle \Rightarrow \sin x = \sin^2 60^{\circ} - \cos^2 60^{\circ}$

$\displaystyle \Rightarrow \sin x = (\frac{\sqrt{3}}{2})^2 - (\frac{1}{2})^2$

$\displaystyle \Rightarrow \sin x = \frac{1}{2} = \sin 30^{\circ}$

$\displaystyle \Rightarrow x = 30^{\circ}$

$\displaystyle \\$

Question 38: Find $\displaystyle x$ if: $\displaystyle \sin x = \sin 60^{\circ} \cos 30^{\circ} + \cos 60^{\circ} \sin 30^{\circ}$

Answer:

$\displaystyle \sin x = \sin 60^{\circ} \cos 30^{\circ} + \cos 60^{\circ} \sin 30^{\circ}$

$\displaystyle \Rightarrow \sin x = \sin 60^{\circ} \cos (90^ - 60^{\circ}) + \cos 60^{\circ} \sin (90^ - 60^{\circ})$

$\displaystyle \Rightarrow \sin x = \sin^2 60^{\circ} + \cos^2 60^{\circ}$

$\displaystyle \Rightarrow \sin x = 1$

$\displaystyle \Rightarrow x = 90^{\circ}$

$\displaystyle \\$

Question 39: Find $\displaystyle x$ if: $\displaystyle \cos x = \cos 60^{\circ} \cos 30^{\circ} - \sin 60^{\circ} \sin 30^{\circ}$

Answer:

$\displaystyle \cos x = \cos 60^{\circ} \cos 30^{\circ} - \sin 60^{\circ} \sin 30^{\circ}$

$\displaystyle \Rightarrow \cos x = \cos 60^{\circ} \cos (90^{\circ} - 60^{\circ}) - \sin 60^{\circ} \sin (90^{\circ} - 60^{\circ})$

$\displaystyle \Rightarrow \cos x = \cos 60^{\circ} \sin 60^{\circ} - \sin 60^{\circ} \cos 30^{\circ}$

$\displaystyle \Rightarrow \cos x = = 0$

$\displaystyle \Rightarrow x = 90^{\circ}$

$\displaystyle \\$

$\displaystyle \text{Question 40: Find the value of } x \text{ if: } \tan x = \frac{\tan 60^{\circ} - \tan 30^{\circ}}{1+ \tan 60^{\circ} \tan 30^{\circ}}$

Answer:

$\displaystyle \tan x = \frac{\tan 60^{\circ} - \tan 30^{\circ}}{1+ \tan 60^{\circ} \tan 30^{\circ}}$

$\displaystyle \Rightarrow \tan x = \frac{\sin 60^{\circ}. \cos 30 - \sin 30^{\circ}. \cos 60}{\cos 60.\cos 30+ \sin 60^{\circ}. \sin 30^{\circ}}$

$\displaystyle \Rightarrow \tan x = \frac{\sin 60^{\circ}. \cos (90^{\circ} - 60^{\circ}) - \sin (90^{\circ} - 60^{\circ}). \cos 60}{\cos 60.\cos 30+ \sin 60^{\circ}. \sin 30^{\circ}}$

$\displaystyle \Rightarrow \tan x = \frac{\sin^2 60^{\circ} - \cos^2 60}{\cos 60.\cos 30+ \sin 60^{\circ}. \sin 30^{\circ}}$

$\displaystyle \Rightarrow \tan x = \frac{(\frac{\sqrt{3}}{2})^2 - (\frac{1}{2})^2}{2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}}$

$\displaystyle \Rightarrow \tan x = \frac{2}{\sqrt{3}}$

Therefore $\displaystyle x = 30^{\circ}$

$\displaystyle \\$

Question 41: Find the value of $\displaystyle x$ if: $\displaystyle \sin 2x = 2 \sin 45^{\circ} \cos 45^{\circ}$

Answer:

$\displaystyle \sin 2x = 2 \sin 45^{\circ} \cos 45^{\circ}$

$\displaystyle \Rightarrow \sin 2x = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$

$\displaystyle \Rightarrow \sin 2x = 1$

$\displaystyle \Rightarrow 2x = 90^{\circ} \Rightarrow x = 45^{\circ}$

$\displaystyle \\$

Question 42: Find the value of $\displaystyle x$ if: $\displaystyle \sin 3x = 2 \sin 30^{\circ} \cos 30^{\circ}$

Answer:

$\displaystyle \sin 3x = 2 \sin 30^{\circ} \cos 30^{\circ}$

$\displaystyle \Rightarrow \sin 3x = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}$

$\displaystyle \Rightarrow \sin 3x = \frac{\sqrt{3}}{2}$

$\displaystyle \Rightarrow 3x = 60^{\circ} \Rightarrow x = 20^{\circ}$

$\displaystyle \\$

Question 43: Find $\displaystyle x$ if: $\displaystyle \cos (2x-6^{\circ}) = \cos^2 30^{\circ} - \cos^2 60^{\circ}$

Answer:

$\displaystyle \cos (2x-6^{\circ}) = \cos^2 30^{\circ} - \cos^2 60^{\circ}$

$\displaystyle \Rightarrow \cos (2x-6^{\circ}) = \sin^2 60^{\circ} - \cos^2 60^{\circ}$

$\displaystyle \Rightarrow \cos (2x-6^{\circ}) = (\frac{\frac{\sqrt{3}}{2}}{2})^2 - (\frac{1}{2})^2$

$\displaystyle \Rightarrow \cos (2x-6^{\circ}) = \frac{1}{2}$

$\displaystyle \Rightarrow 2x - 6 = 60 \Rightarrow x = 33^{\circ}$

$\displaystyle \\$

Question 44: Find the value of $\displaystyle A$ where $\displaystyle 0^{\circ} \leq A \leq 90^{\circ} \text{ : } \sin (90^{\circ} - 3A) . \mathrm{cosec} 42^{\circ} = 1$

Answer:

$\displaystyle \sin (90^{\circ} - 3A) . \mathrm{cosec} 42^{\circ} = 1$

$\displaystyle \Rightarrow \sin (90^{\circ} - 3A) = \sin 42^{\circ}$

$\displaystyle \Rightarrow 90^{\circ} - 3A = 42^{\circ} \Rightarrow A = 16^{\circ}$

$\displaystyle \\$

Question 45: Find the value of $\displaystyle A$ where $\displaystyle 0^{\circ} \leq A \leq 90^{\circ} \text{ : } \cos (90^{\circ} - A) . \sec 77^{\circ} = 1$

Answer:

$\displaystyle \cos (90^{\circ} - A) . \sec 77^{\circ} = 1$

$\displaystyle \Rightarrow \cos (90^{\circ} - A) = \cos 77^{\circ}$

$\displaystyle \Rightarrow (90^{\circ} - A) = 77^{\circ} \Rightarrow A = 13^{\circ}$

$\displaystyle \\$

$\displaystyle \text{Question 46: Prove that: } \frac{\cos (90^{\circ}-A).\cos A}{\cot A} = 1 - \cos^2 A$

Answer:

$\displaystyle \text{LHS } = \frac{\cos (90^{\circ}-A).\cos A}{\cot A}$

$\displaystyle = \frac{\sin A.\cos A . \sin A}{\cos A}$

$\displaystyle = \sin^2 A = 1- \cos^2 A = \text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 47: Prove that: } \frac{\sin A . \sin (90^{\circ}-A)}{\tan (90^{\circ}-A)} = 1 - \sin^2 A$

Answer:

$\displaystyle \text{LHS } = \frac{\sin A. \sin (90^{\circ}-A)}{\tan A}$

$\displaystyle = \frac{\sin A.\cos A . \cos A}{\sin A}$

$\displaystyle = \cos^2 A = 1- \sin^2 A = \text{ RHS. Hence proved. }$

$\displaystyle \\$

$\displaystyle \text{Question 48: Evaluate: } \frac{\sin 35^{\circ}.\cos 55^{\circ} + \cos 35^{\circ}. \sin 55^{\circ}}{\mathrm{cosec}^2 10^{\circ} - \tan^2 80^{\circ}}$ [2010]

Answer:

$\displaystyle \frac{\sin 35^{\circ}.\cos 55^{\circ} + \cos 35^{\circ}. \sin 55^{\circ}}{\mathrm{cosec}^2 10^{\circ} - \tan^2 80^{\circ}}$

$\displaystyle = \frac{\sin 35^{\circ}.\cos (90^{\circ} - 35^{\circ}) + \cos 35^{\circ}. \sin (90^{\circ} - 35^{\circ})}{\mathrm{cosec}^2 10^{\circ} - \tan^2 (90^{\circ} - 10^{\circ}) }$