Prove the following identities,
[question] Question. Prove the following identities, where the angles involved are acute angles for which the following expressions are defined. (i) $(\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}$. (ii) $\frac{\cos \mathbf{A}}{\mathbf{1}+\sin \mathbf{A}}+\frac{\mathbf{1}+\sin \mathbf{A}}{\cos \mathbf{A}}=2 \sec \mathrm{A}$. (iii) $\frac{\tan \theta}{\mathbf{1}-\cot \theta}+\frac{\cot \theta}{\mathbf{1}-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$ (i...
Choose the correct option. Justify your choice :
[question] Question. 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) $(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=$ (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) $\frac{\mathbf{1}+\tan ^{2} \mathbf{A}}{\mathbf{1}+\cot ^{2} \mathbf{A}}=$ (A) $\sec ^{2} \mathrm{~A}$ (B) $-1$ (C) $\cot ^{2} A$ (D) $\tan ^{2} \mathrm{~A}$ [/question] [solution] Sol...
Write all the other trigonometric ratios of $\angle \mathrm{A}$ in terms of $\sec \mathrm{A}$.
[question] Question. Write all the other trigonometric ratios of $\angle \mathrm{A}$ in terms of $\sec \mathrm{A}$. [/question] [solution] Solution: (i) $\sin A=\sqrt{1-\cos ^{2} A}$ $=\sqrt{1-\frac{1}{\sec ^{2} A}}=\frac{\sqrt{\sec ^{2} A-1}}{\sec A}$ (ii) $\cos A=\frac{1}{\sec A}$ (iii) $\tan A=\sqrt{\sec ^{2} \mathbf{A}-\mathbf{1}}$ (iv) $\cot A=\frac{1}{\tan A}=\frac{1}{\sqrt{\sec ^{2} A-1}}$ (v) $\operatorname{cosec} A=\frac{1}{\sin A}=\frac{\sec A}{\sqrt{\sec ^{2} A-1}}$ [/solution]...
Express the trigonometric ratios sin A,
[question] Question. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. [/question] [solution] Solution: We have $\operatorname{cosec}^{2} A-\cot ^{2} A=1$ $\Rightarrow \operatorname{cosec}^{2} A=1+\cot ^{2} A$ $\Rightarrow(\operatorname{cosec} A)^{2}=\cot ^{2} A+1$ $\Rightarrow\left(\frac{1}{\sin A}\right)^{2}=\cot ^{2} A+1$ $\Rightarrow(\sin \mathrm{A})^{2}=\frac{\mathbf{1}}{\cot ^{2} \mathbf{A}+\mathbf{1}}$ $\Rightarrow \sin A=\pm \frac{1}{\sqrt{\cot ^{2} A+1}}$ We rej...
Express $\sin 67^{\circ}+\cos 75^{\circ}$
[question] Question. Express $\sin 67^{\circ}+\cos 75^{\circ}$ in terms of trigonometric ratios of angles between $0^{\circ}$ and $45^{\circ}$. [/question] [solution] Solution: sin 67° + cos 75° = sin(90° – 23°) + cos (90° – 15°) = cos 23° + sin 15° [/solution]...
If A, B and C are interior angles of a triangle ABC,
[question] Question. If $A, B$ and $C$ are interior angles of a triangle $A B C$, then show that $\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2} .$ [/question] [solution] Solution: A + B + C = 180° $\Rightarrow \mathrm{B}+\mathrm{C}=180^{\circ}-\mathrm{A}$ $\Rightarrow \frac{\mathbf{B}+\mathbf{C}}{2}=\frac{180^{\circ}-\mathbf{A}}{2} \Rightarrow \frac{\mathbf{B}+\mathbf{C}}{2}=\left(\mathbf{M}^{\circ}-\frac{\mathbf{A}}{2}\right)$ $\Rightarrow \sin \left(\frac{\mathbf{B}+\mathbf{C}}{\mathbf{2}}\...
If sec 4 A = cosec (A – 20°),
[question] Question. If sec 4 A = cosec (A – 20°), where 4 A is an acute angle, find the value of A. [/question] [solution] Solution: sec 4 A = cosec (A – 20°) $\Rightarrow \operatorname{cosec}\left(90^{\circ}-4 \mathrm{~A}\right)=\operatorname{cosec}\left(\mathrm{A}-20^{\circ}\right)$ $\left\{\because \operatorname{cosec}\left(90^{\circ}-\theta\right)=\sec \theta\right\}$ $\Rightarrow 90^{\circ}-4 \mathrm{~A}=\mathrm{A}-20^{\circ}$ $\Rightarrow 5 \mathrm{~A}=110^{\circ} \Rightarrow \mathrm{A}=2...
If tan A = cot B, prove that A + B = 90°.
[question] Question. If tan A = cot B, prove that A + B = 90°. [/question] [solution] Solution: tan A = cot B tan A = tan (90°–B) $\therefore \mathrm{A}=90^{\circ}-\mathrm{B}$ A + B = 90° [/solution]...
If tan 2 A = cot (A – 18°), where 2 A is an acute angle,
[question] Question. If tan 2 A = cot (A – 18°), where 2 A is an acute angle, find the value of A. [/question] [solution] Solution: tan2A = cot(A – 18°) cot (90° – 2A) = cot (A – 18°) 90° – 2A = A – 18° 108° = 3A A = 36° [/solution]...
Show that
[question] Question. Show that (i) tan 48° tan 23° tan 42° tan 67° = 1 (ii) cos 38° cos 52° – sin 38° sin 52° = 0. [/question] [solution] Solution: (i) $\mathrm{LHS}=\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}$ $=\tan 48^{\circ} \times \tan 23^{\circ} \times \tan \left(90^{\circ}-48^{\circ}\right)$ $\times \tan \left(90^{\circ}-23^{\circ}\right)$ $=\tan 48^{\circ} \times \tan 23^{\circ} \times \cot 48^{\circ} \times \cot 23^{\circ}$ $=\tan 48^{\circ} \times \tan 23^{\circ} \t...
Evaluate :
[question] Question. Evaluate : (i) $\frac{\sin 18^{\circ}}{\cos 72^{\circ}}$ (ii) $\frac{\tan 26^{\circ}}{\cot 64^{\circ}}$ (iii) $\cos 48^{\circ}-\sin 42^{\circ}$ (iv) $\operatorname{cosec} 31^{\circ}-\sec 59^{\circ}$ [/question] [solution] Solution: (i) $\frac{\sin 18^{\circ}}{\cos 72^{\circ}}=\frac{\sin 18^{\circ}}{\cos \left(90^{\circ}-18^{\circ}\right)}=\frac{\sin 18^{\circ}}{\sin 18^{\circ}}=1$. $\left\{\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right\}$ (ii) $\frac{\tan 26^...
State whether the following are true or false. Justify your answer
[question] Question. 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) $\cot \mathrm{A}$ is not defined for $\mathrm{A}=0^{\circ}$. [/question] [solution] Solution: (i) False. When A = 60°, B = 30° LHS = sin (A + B) = sin (60° + 30°) = sin 90° = 1 RHS = si...
If $\tan (\mathrm{A}+\mathrm{B})=\sqrt{3}$
[question] Question. If $\tan (\mathrm{A}+\mathrm{B})=\sqrt{3}$ and $\tan (\mathrm{A}-\mathrm{B})=\frac{\mathbf{1}}{\sqrt{\mathbf{3}}}$ $0^{\circ}\mathrm{B}$, find $\mathrm{A}$ and $\mathrm{B}$. [/question] [solution] Solution: $\tan (\mathrm{A}+\mathrm{B})=\sqrt{\mathbf{3}} \quad \Rightarrow \mathrm{A}+\mathrm{B}=60^{\circ}$ ...(1) $\tan (\mathrm{A}-\mathrm{B})=\frac{1}{\sqrt{3}} \quad \Rightarrow \mathrm{A}-\mathrm{B}=30^{\circ}$ ...(2) Adding (1) and (2), $2 \mathrm{~A}=90^{\circ} \Rightarrow...
Evaluate :
[question] Question. Evaluate : (i) $\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}$ (ii) $2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}$ (iii) $\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$ (iv) $\frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}$ (v) $\frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ...
State whether the following are true or false.
[question] Question. State whether the following are true or false. Justify your answer (i) The value of tan A is always less than 1. (ii) $\sec A=\frac{12}{5}$ 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) $\sin \theta=\frac{4}{3}$ for some angle $\theta$. [/question] [solution] Solution: (i) False. We know that $60^{\circ}=\sqrt{\mathbf{3}}>\mathbf{1}$. (ii) True. We know that value of $\sec A$ is alwa...
In triangle ABC right angled at B
[question] Question. In triangle $\mathrm{ABC}$ right angled at $\mathrm{B}$, if $\tan \mathrm{A}=\frac{\mathbf{1}}{\sqrt{\mathbf{3}}}$, find the value of : (i) $\sin A \cos C+\cos A \sin C$ (ii) $\cos \mathrm{A} \cos \mathrm{C}-\sin \mathrm{A} \sin \mathrm{C}$. [/question] [solution] Solution: $\tan A=\frac{1}{\sqrt{3}}$ $\frac{\mathbf{B C}}{\mathbf{B A}}=\frac{\mathbf{1}}{\sqrt{\mathbf{3}}}$ $\mathrm{BC}=\mathrm{k}$ and $\mathrm{BA}=\sqrt{\mathbf{3} \mathbf{k}}$ $\mathrm{AC}^{2}=\mathrm{BC}^{2...
If $3 \cot \mathrm{A}=4$, check whether
[question] Question. If $3 \cot \mathrm{A}=4$, check whether $\frac{1-\tan ^{2} \mathbf{A}}{1+\tan ^{2} \mathbf{A}}=\cos ^{2} \mathrm{~A}-\sin ^{2} \mathrm{~A}$ or not. [/question] [solution] Solution: In figure, $3 \cot A=4$ $\Rightarrow \cot A=\frac{4}{3}$ $\Rightarrow \frac{A B}{B C}=\frac{4}{3}$ $\Rightarrow \mathrm{AB}=4 \mathrm{k}$ and $\mathrm{BC}=3 \mathrm{k}$ Then $\sin \mathrm{A}=\frac{\mathbf{B C}}{\mathbf{A C}}=\frac{\mathbf{3 k}}{\mathbf{5 k}}=\frac{\mathbf{3}}{\mathbf{5}}$, $\cos A...
If $\cot \theta=\frac{7}{8}$, evaluate :
[question] Question. If $\cot \theta=\frac{7}{8}$, evaluate : (i) $\frac{(\mathbf{1}+\sin \theta)(\mathbf{1}-\sin \theta)}{\mathbf{( 1}+\cos \theta)(\mathbf{1}-\cos \theta)}$ (ii) $\cot ^{2} \theta$ [/question] [solution] Solution: In figure, $\cot \theta=\frac{\mathbf{7}}{\mathbf{8}}$ $\Rightarrow \quad \frac{\mathbf{A B}}{\mathbf{B C}}=\frac{\mathbf{7}}{\mathbf{8}}$ $\Rightarrow \quad \mathrm{AB}=7 \mathrm{k}$ and $B C=8 \mathrm{k}$ Now, $A C^{2}=A B^{2}+B C^{2}=(7 k)^{2}+(8 k)^{2}$ $=113 \mat...
If $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are acute angles such that $\cos \mathrm{A}=\cos \mathrm{B}$,
[question] Question. If $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are acute angles such that $\cos \mathrm{A}=\cos \mathrm{B}$, then show that $\angle \mathrm{A}=\angle \mathrm{B}$. [/question] [solution] Solution: In figure $\angle \mathrm{A}$ and $\angle \mathrm{B}$ are acute angles of $\triangle \mathrm{ABC}$. Draw $C D \perp A B$ We are given that $\cos A=\cos B$ $\Rightarrow \frac{\mathbf{A D}}{\mathbf{A C}}=\frac{\mathbf{B D}}{\mathbf{B C}}$ $\Rightarrow \frac{\mathbf{A D}}{\mathbf{B D}}...
Given $\sec \theta=\frac{\mathbf{1 3}}{\mathbf{1 2}}$, calculate all other trigonometric ratios.
[question] Question. Given $\sec \theta=\frac{\mathbf{1 3}}{\mathbf{1 2}}$, calculate all other trigonometric ratios. [/question] [solution] Solution: $\sec \theta=\frac{13}{12}$ $\Rightarrow \frac{A C}{B C}=\frac{13}{12}$ By Pythagoras Theorem, $\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$ $(13 k)^{2}=A B^{2}+(12 k)^{2}$ $\mathrm{AB}^{2}=169 \mathrm{k}^{2}-144 \mathrm{k}^{2}$ $\mathrm{AB}=\sqrt{25 \mathrm{k}^{2}}=5 \mathrm{k}$ $\sin \theta=\frac{\mathbf{A B}}{\mathbf{A C}}=\frac{5 \mathbf{k...
Given 15 cot A = 8, find sin A and sec A.
[question] Question. Given 15 cot A = 8, find sin A and sec A. [/question] [solution] Solution: $\cot A=\frac{8}{15}$ $\Rightarrow \frac{\mathbf{A B}}{\mathbf{B C}}=\frac{\mathbf{8}}{\mathbf{1 5}}$ $\Rightarrow \mathrm{AB}=8 \mathrm{k}$ and $\mathrm{BC}=15 \mathrm{k}$ Now, $A C=\sqrt{(8 k)^{2}+(15 k)^{2}}=17 k$ $\sin A=\frac{B C}{A C}=\frac{15 k}{17 k}=\frac{15}{17}, \quad \sec A=\frac{A C}{A B}=\frac{17 k}{8 k}=\frac{17}{8}$ [/solution]...
If $\sin A=\frac{\mathbf{3}}{\mathbf{4}}$, calculate $\cos A$ and $\tan A$.
[question] Question. If $\sin A=\frac{\mathbf{3}}{\mathbf{4}}$, calculate $\cos A$ and $\tan A$. [/question] [solution] Solution: In figure, $\sin A=\frac{3}{4}$ $\Rightarrow \frac{B C}{A C}=\frac{3}{4}$ $\Rightarrow \mathrm{BC}=3 \mathrm{k}$ and $A C=4 k$ where k is the constant of proportionality. By Pythagoras Theorem, $\mathrm{AB}^{2}=\mathrm{AC}^{2}-\mathrm{BC}^{2}=(4 \mathrm{k})^{2}-(3 \mathrm{k})^{2}=7 \mathrm{k}^{2}$ $\Rightarrow \mathrm{AB}=\sqrt{7} \mathrm{k}$ So, $\cos A=\frac{A B}{A ...
In fig, find tan P – cot R.
[question] Question. In fig, find tan P – cot R. [/question] [solution] Solution: In figure, by the Pythagoras Theorem $\mathrm{QR}^{2}=\mathrm{PR}^{2}-\mathrm{PQ}^{2}=(13)^{2}-(12)^{2}=25$ $\Rightarrow \mathrm{QR}=\sqrt{25}=5 \mathrm{~cm}$ In $\triangle \mathrm{PQR}$ right angled at $\mathrm{Q}, \mathrm{QR}=5 \mathrm{~cm}$ is side opposite to the angle $\mathrm{P}$ and $\mathrm{PQ}=12 \mathrm{~cm}$ is side adjacent to the angle P. Therefore, $\tan \mathrm{P}=\frac{\mathbf{Q R}}{\mathbf{P Q}}=\f...
In $\triangle \mathrm{ABC}$, right angled at $\mathrm{B}, \mathrm{AB}=24 \mathrm{~cm}$,$B C=7 \mathrm{~cm}$. Determine :
[question] Question. In $\triangle \mathrm{ABC}$, right angled at $\mathrm{B}, \mathrm{AB}=24 \mathrm{~cm}$,$B C=7 \mathrm{~cm}$. Determine : (i) $\sin A, \cos A$ (ii) $\sin C, \cos C$. [/question] [solution] Solution: By Pythagoras Theorem, $\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}=(24)^{2}+(7)^{2}=625$ $\Rightarrow \mathrm{AC}=\sqrt{625}=25 \mathrm{~cm}$ (i) $\sin \mathrm{A}=\frac{\mathbf{B C}}{\mathbf{A C}}\left\{\right.$ ie, $\left.\frac{\text { sile opposite to ang } \mathbf{A}}{\tex...