Question 512 Marks
Prove.
$\sin ^4 A-\cos ^4 A=2 \sin ^2 A-1$
Answer$\text { LHS }=\sin ^4 A-\cos ^4 A $
$ =\left(\sin ^2 A\right)^2-\left(\cos ^2 A\right)^2$
$ =\left(\sin ^2 A+\cos ^2 A\right)\left(\sin ^2 A-\cos ^2 A\right)$
$ =\sin ^2 A-\cos ^2 A $
$=\sin ^2 A-\left(1-\sin ^2 A\right) $
$=2 \sin ^2 A-1 \text { RHS }$
View full question & answer→Question 522 Marks
Prove.
$\tan A-\cot A=\frac{1-2 \cos ^2 A}{\sin A \cos A}$
Answer$\tan A-\cot A=\frac{\sin A}{\cos A}-\frac{\cos A}{\sin A}$
$=\frac{\sin ^2 A-\cos ^2 A}{\sin A \cos A}$
$=\frac{1-\cos ^2 A-\cos ^2 A}{\sin A \cos A}\left(\because \sin ^2 A =1-\cos ^2 A \right) $
$ =\frac{1-2 \cos 2}{\sin A \cos A}$
View full question & answer→Question 532 Marks
Prove.
$\frac{1}{\sin A+\cos A}+\frac{1}{\sin A-\cos A}=\frac{2 \sin A}{1-2 \cos ^2 A}$
Answer$\text { LHS }=\frac{1}{\sin A+\cos A}+\frac{1}{\sin A-\cos A} $
$=\frac{\sin A-\cos A+\sin A+\cos A}{\sin ^2 A-\cos ^2 A} $
$ =\frac{2 \sin A}{1-\cos ^2 A-\cos ^2 A}=\frac{2 \sin A}{1-2 \cos ^2 A}$
View full question & answer→Question 542 Marks
Prove.
$1-\frac{\cos ^2 A}{1+\sin A}=\sin A$
Answer$\text { LHS }=1-\frac{\cos ^2 A}{1+\sin A}$
$ =\frac{1+\sin A-\cos ^2 A}{1+\sin A}$
$=\frac{\sin A+\sin ^2 A}{1+\sin ^2 A}$
$ =\frac{\sin A(1+\sin A)}{1+\sin A} $
$ =\sin A=\text { RHS }$
View full question & answer→Question 552 Marks
Prove.
$\sqrt{\frac{1-\sin A}{1+\sin A}}=\frac{\cos A}{1+\sin A}$
Answer$\text { LHS }=\sqrt{\frac{1-\sin A}{1+\sin A}} $
$=\sqrt{\frac{1-\sin A}{1+\sin A} \times \frac{1+\sin A}{1+\sin A}}$
$ =\sqrt{\frac{1-\sin ^2 A}{(1+\sin A)^2}} $
$ =\sqrt{\frac{\cos ^2 A}{(1+\sin A)^2}} $
$\frac{\cos A}{1+\sin A}=\text { RHS }$
View full question & answer→Question 562 Marks
Prove.$\sqrt{\frac{1-\cos A}{1+\cos A}}=\frac{\sin A}{(1+\cos A)}$
Answer$\text { LHS }=\sqrt{\frac{1-\cos A}{1+\cos A}} $
$=\sqrt{\frac{1-\cos A}{1+\cos A} \times \frac{1+\cos A}{1+\cos A}}$
$ =\sqrt{\frac{1-\cos ^2 A}{(1+\cos A)^2}} $
$=\sqrt{\frac{\sin ^2 A}{(1+\cos A)^2}} $
$=\frac{\sin A}{1+\cos A}=\text { RHS }$
View full question & answer→Question 572 Marks
Prove.
$\sqrt{\frac{1-\cos A}{1+\cos A}}=\operatorname{cosec} A-\cot A$
Answer$\text { LHS }=\sqrt{\frac{1-\cos A}{1+\cos A}}$
$ =\sqrt{\frac{1-\cos A}{1+\cos A} \times \frac{1-\cos A}{1-\cos A}} $
$ =\sqrt{\frac{(1+\cos A)^2}{1-\cos ^2 A}} \\ =\sqrt{\frac{(1+\cos A)^2}{\sin ^2 A}} $
$ =\frac{1-\cos A}{\sin A} $
$=\operatorname{cosec} A-\cot A=\text { RHS }$
View full question & answer→Question 582 Marks
Prove.
$\frac{1}{\tan A+\cot A}=\cos A \sin A$
Answer$\frac{1}{\tan A+\cot A}=\sin A \cos A $
$ \text { LHS }=\frac{1}{\tan A+\operatorname{Cot} A}$
$ =\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}=\frac{1}{\frac{\sin ^2 A+\cos ^2}{\sin A \cos A}}$
$\frac{1}{\frac{1}{\sin A \cos A}}\left(\because \sin ^2 A +\cos ^2 A =1\right) $
$=\sin A \cos A =\text { RHS }$
View full question & answer→Question 592 Marks
Prove.
$\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$
Answer$\text { LHS }=\sqrt{\frac{1+\sin A}{1-\sin A}} $
$ =\sqrt{\frac{1+\sin A}{1-\sin A} \times \frac{1+\sin A}{1+\sin A}}$
$=\sqrt{\frac{(1+\sin A)^2}{1-\sin ^2 A}}=\sqrt{\frac{(1+\sin A)^2}{\cos ^2 A}} $
$=\frac{1+\sin A}{\cos A} $
$=\sec A+\tan A=\text { RHS }$
View full question & answer→Question 602 Marks
Prove.
$\frac{1+\sin A}{1-\sin A}=\frac{c o \sec A+1}{c o \sin A-1}$
Answer$ LHS =\frac{1+\sin A}{1-\sin A} $
$RHS =\frac{\operatorname{cosec} A+1}{\operatorname{cosec} A-1}=\frac{\frac{1}{\sin A}+1}{\frac{1}{\sin A}-1} $
$=\frac{1+\sin A}{1-\sin A}$
View full question & answer→Question 612 Marks
Prove.
$\frac{\sec A}{\sec A+1}+\frac{\sec A}{\sec A-1}=2 \operatorname{cosec} c^2 A$
Answer$\text { LHS }=\frac{\sec A}{\sec A+1}+\frac{\sec A}{\sec A-1} $
$ =\frac{\sec ^2 A-\sec A+\sec ^2 A+\sec A}{\sec ^2 A-1} $
$=\frac{2 \sec ^2 A}{\tan ^2 A}\left(\because \sec ^2 A-1=\tan ^2 A\right)$
$\frac{\frac{2}{\cos ^2 A}}{\frac{\sin ^2 A}{\cos ^2 A}}=\frac{2}{\sin ^2 A}=2 \operatorname{cosec^2} A=\text { RHS }$
View full question & answer→Question 622 Marks
Prove.
$\frac{1}{1-\sin A}+\frac{1}{1+\sin A}=2 \sec ^2 A$
Answer$\text { LHS }=\frac{1}{1-\sin A}+\frac{1}{1+\sin A}$
$ =\frac{1+\sin A+1-\sin A}{(1-\sin A)(1+\sin A)}$
$ =\frac{2}{1-\sin ^2 A} $
$=\frac{2}{\cos ^2 A}$
$ =2 \sec ^2 A=\text { RHS }$
View full question & answer→Question 632 Marks
Prove.
$\frac{1}{1+\cos A}+\frac{1}{1-\cos A}=2 \operatorname{cosec}{ }^2 A$
Answer$\text { LHS }=\frac{1}{1+\cos A}+\frac{1}{1-\cos A} $
$ =\frac{1-\cos A+1+\cos A}{(1+\cos A)(1-\cos A)} $
$ =\frac{2}{1-\cos ^2 A} $
$=\frac{2}{\sin ^2 A} $
$=2 \operatorname{cosec}^2 A =\text { RHS }$
View full question & answer→Question 642 Marks
prove.
$\sec^2A\ cosec^2A = \tan^2A + \cot^2A + 2$
Answer$\sec ^2 A \operatorname{cosec}^2 A$
$=\left(1+\tan ^2 A\right)\left(1+\cot ^2 A\right)$
$=1+\cot ^2 A+\tan ^2 A+\tan ^2 A \cdot \cot ^2 A$
$=1+\cot ^2 A+\tan ^2 A+\frac{1}{\cot ^2 A} \times \cot ^2 A$
$=\tan ^2 A+\cot ^2 A+2$
View full question & answer→Question 652 Marks
Prove.
$\operatorname{cosec} A+\cot A=\frac{1}{\operatorname{cosec} A-\cot A}$
Answer$\text { LHS }=\operatorname{cosec} A+\cot A$
$ =\frac{\operatorname{cosec} A+\cot A}{1} \times \frac{\operatorname{cosec} A-\cot A}{\operatorname{cosec} A-\cot A} $
$ =\frac{\operatorname{cosec} c^2 A-\cot ^2 A}{\operatorname{cosec} A-\cot A}=\frac{1+\cot ^2 A-\cot ^2 A}{\operatorname{cosec} A-\cot A} $
$=\frac{1}{\operatorname{cosec} A-\cot A}=\text { RHS }$
View full question & answer→Question 662 Marks
Prove.
$(cosA + sinA)^2 + (cosA - sinA)^2 = 2$
Answer$LHS =(cosA + sinA)^2 + (cosA - sinA)^2$
$= \cos^2A + \sin^2A + 2cos A.\sin A + \cos^2A + \sin^2A - 2cos A.\sin A$
$= 2(\cos^2A + \sin^2A) = 2 = RHS$
View full question & answer→Question 672 Marks
Prove.
$(sec A - \cos A) (sec A + \cos A) = \sin^2 A + \tan^2 A$
Answer$LHS =(sec A - \cos A) (sec A + \cos A)$
$= sec^2 A - \cos^2 A$
$= (1 + \tan^2 A) - (1 - \sin^2 A)$
$= \sin^2 A + \tan^2 A = RHS$
View full question & answer→Question 682 Marks
Prove.
$(cosec A + \sin A) (cosec A - \sin A) = \cot^2 A + \cos^2 A$
Answer$LHS =(cosec A + \sin A) (cosec A - \sin A)$
$= cosec^2 A - \sin^2 A$
$= (1 + \cot^2 A) - (1 - \cos^2 A)$
$=\cot^2 A + \cos^2 A = RHS$
View full question & answer→Question 692 Marks
Prove.
$\cot^2 A - \cos^2 A = \cos^2 A.\cot^2 A$
Answer$\text { LHS }=\cot ^2 A-\cos ^2 A$
$=\frac{\cos ^2 A}{\sin ^2 A}-\cos ^2 A=\frac{\cos ^2 A\left(1-\sin ^2 A\right)}{\sin ^2 A}$
$=\cos ^2 A \frac{\cos ^2 A}{\sin ^2 A}=\cos ^2 A \cdot \cot ^2 A=\text { RHS }$
View full question & answer→Question 702 Marks
Prove.
$\tan^2A - \sin^2A = \tan^2A \sin^2A$
Answer$\text { LHS }=\tan ^2 A-\sin ^2 A$
$=\frac{\sin ^2 A}{\cos ^2 A}-\sin ^2 A=\frac{\sin ^2 A\left(1-\cos ^2 A\right)}{\cos ^2 A}$
$=\frac{\sin ^2 A}{\cos ^2 A} \cdot \sin ^2 A=\tan ^2 A \cdot \sin ^2 A=\text { RHS }$
View full question & answer→Question 712 Marks
Prove.
$\frac{\left(1+\tan ^2 A\right) \cot A}{\operatorname{cosec}^2 A}=\tan A$
Answer$\text { LHS }=\frac{\left(1+\tan ^2 A\right) \cot A}{\operatorname{cosec} A} $
$\frac{\sec ^2 A \cot A}{\cos e c^2 A}\left(\because \sec ^2 A =1+\tan ^2 A \right) $
$\frac{\frac{1}{\cos ^2 A} \times \frac{\cos A}{\sin A}}{\frac{1}{\sin ^2 A}}=\frac{\frac{1}{\cos A \sin A}}{\frac{1}{\sin ^2 A}}$
$=\frac{\sin A}{\cos A}=\tan A=\text { RHS }$
View full question & answer→Question 722 Marks
Prove.
$\sec^2 A + cosec^2 A = \sec^2 A cosec^2 A$
Answer$\text { LHS }=\sec ^2 A +\operatorname{cosec}^2 A$
$=\frac{1}{\cos ^2 A}+\frac{1}{\sin ^2 A}=\frac{\sin ^2 A+\cos ^2 A}{\cos ^2 A \sin ^2 A}$
$=\frac{1}{\cos ^2 A \sin ^2 A}=\sec ^2 A \operatorname{cosec} 2=\text { RHS }$
View full question & answer→Question 732 Marks
If $A$ and $B$ are complementary angles, prove that: $\cot A \cot B - \sin A \cos B - \cos A \sin B = 0$
AnswerSince, $A$ and $B$ are complementary angles, $A + B = 90^\circ$
$\cot A \cot B - \sin A \cos B - \cos A \sin B$
$= \cot A \cot(90^\circ - A) - \sin A \cos(90^\circ - A) - \cos A \sin(90^\circ - A)$
$= \cot A \tan A - \sin A \sin A - \cos A \cos A$
$= 1 - (\sin^2A + \cos^2A)$
$= 1 - 1$
$= 0$
View full question & answer→Question 742 Marks
Evaluate $\frac{\cos 75^{\circ}}{\sin 15^{\circ}}+\frac{\sin 12^{\circ}}{\cos 78^{\circ}}-\frac{\cos 18^{\circ}}{\sin 72^{\circ}}$
Answer$\frac{\cos 75^{\circ}}{\sin 15^{\circ}}+\frac{\sin 12^{\circ}}{\cos 78^{\circ}}-\frac{\cos 18^{\circ}}{\sin 72^{\circ}}$
$=\frac{\cos \left(90^{\circ}-15^{\circ}\right)}{\sin 15^{\circ}}+\frac{\sin \left(90^{\circ}-78^{\circ}\right)}{\cos 78^{\circ}}-\frac{\cos \left(90^{\circ}-72^{\circ}\right)}{\sin 72^{\circ}}$
$=\frac{\sin 15^{\circ}}{\sin 15^{\circ}}+\frac{\cos 78^{\circ}}{\cos 78^{\circ}}-\frac{\sin 72^{\circ}}{\sin 72^{\circ}}$
$=1+1-1$
$=1$
View full question & answer→Question 752 Marks
Evaluate $\frac{3 \sin 72^{\circ}}{\cos 18^{\circ}}-\frac{\sec 32^{\circ}}{\operatorname{cosec} 58^{\circ}}$
Answer$\frac{3 \sin 72^{\circ}}{\cos 18^{\circ}}-\frac{\sec 32^{\circ}}{\cos e c 58^{\circ}}$
$=\frac{3 \sin \left(90^{\circ}-18^{\circ}\right)}{\cos 18^{\circ}}-\frac{\sec \left(90^{\circ}-58^{\circ}\right)}{\cos e c 58^{\circ}}$
$=\frac{3 \cos 18^{\circ}}{\cos 18^{\circ}}-\frac{\operatorname{cosec} 58^{\circ}}{\operatorname{cosec} 58^{\circ}}$
$=3-1$
$=2$
View full question & answer→Question 762 Marks
Evaluate $\frac{5 \sin 66^{\circ}}{\cos 24^{\circ}}-\frac{2 \cot 85^{\circ}}{\tan 5^{\circ}}$
Answer$\frac{5 \sin 66^{\circ}}{\cos 24^{\circ}}-\frac{2 \cot 85^{\circ}}{\tan 5^{\circ}}$
$=\frac{5 \sin \left(90^{\circ}-24^{\circ}\right)}{\cos 24^{\circ}}-\frac{2 \cot \left(90^{\circ}-5^{\circ}\right)}{\tan 5^{\circ}}$
$=\frac{5 \cos 24^{\circ}}{\cos 24^{\circ}}-\frac{2 \tan 5^{\circ}}{\tan 5^{\circ}}$
$=5-2$
$=3$
View full question & answer→Question 772 Marks
Evaluate $\sec 26^{\circ} \sin 64^{\circ}+\frac{\cos e c 33^{\circ}}{\sec 57^{\circ}}$
Answer$\sec 26^{\circ} \sin 64^{\circ}+\frac{\cos e c 33^{\circ}}{\sec 57^{\circ}}$
$=\sec \left(90^{\circ}-64^{\circ}\right) \sin 64^{\circ}+\frac{\operatorname{cosec}\left(90^{\circ}-57^{\circ}\right)}{\sec 57^{\circ}}$
$=\cos e c 64^{\circ} \sin 64^{\circ}+\frac{\sec 57^{\circ}}{\sec 57^{\circ}}$
$=1+1$
$=2$
View full question & answer→Question 782 Marks
If $x=a \cos \theta$ and $y=b \cot \theta$, show that: $\frac{a^2}{x^2}-\frac{b^2}{y^2}=1$
Answer$\frac{a^2}{x^2}-\frac{b^2}{y^2}$
$=\frac{a^2}{a^2 \cos ^2 \theta}-\frac{b^2}{b^2 \cot ^2 \theta}$
$=\frac{1}{\cos ^2 \theta}-\frac{\sin ^2 \theta}{\cos ^2 \theta}$
$=\frac{1-\sin ^2 \theta}{\cos ^2 \theta}$
$=\frac{\cos ^2 \theta}{\cos ^2 \theta}$
$=1$
View full question & answer→Question 792 Marks
If $sinA + cosA = p$ and $secA + cosecA = q,$ then prove that: $q(p^2 - 1) = 2p$
Answer$q(p^2 - 1) = (secA + cosecA) [(sinA + cosA)^2 - 1]$
$= (secA + cosecA) [(\sin^2A + \cos^2A + 2sinA \ cosA) - 1]$
$= (secA + cosecA) [(1 + 2sinA \ cosA)-1]$
$= (secA + cosecA) (2sinA \ cosA)$
$= 2sinA + 2cosA$
$= 2p$
View full question & answer→Question 802 Marks
Find $A,$ if $0^\circ \leq A \leq 90^\circ$ and $\cos^2A - cosA = 0$
Answer$\cos^2A - cosA = 0$
$\Rightarrow cosA (cosA - 1) = 0$
$\Rightarrow cosA = 0$ or $cosA = 1$
We know $\cos 90^\circ = 0$ and $\cos 0^\circ = 1$
Hence, $A = 90^\circ$ or $0^\circ$
View full question & answer→Question 812 Marks
Find $A,$ if $0^\circ \leq A \leq 90^\circ$ and $4sin^2A - 3 = 0$
Answer$4 \sin ^2 A -3=0$
$\Rightarrow \sin ^2 A=\frac{3}{4}$
$\Rightarrow \sin A=\frac{\sqrt{3}}{2}$
We know $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$
Hence, $A=60^{\circ}$
View full question & answer→Question 822 Marks
Find A, if 0° ≤ A ≤ 90° and sin 3A - 1 = 0
Answersin 3A - 1 = 0
⇒ sin3A = 1
We know sin 90° = 1
∴ 3A = 90°
Hence, A = 30°
View full question & answer→Question 832 Marks
Find $A,$ if $0^\circ \leq A \leq 90^\circ$ and $2cos^2A - 1 = 0$
Answer$2cos^2A - 1 = 0$
$\Rightarrow \cos ^2 A=\frac{1}{2}$
$\Rightarrow \cos A=\frac{1}{\sqrt{2}}$
We know $\cos 45^{\circ}=\frac{1}{\sqrt{2}}$
Hence, $A=45^{\circ}$
View full question & answer→Question 842 Marks
Prove the following identitie : $\sqrt{\frac{1-\cos A}{1+\cos A}}=\frac{\sin A}{1+\cos A}$
Answer$\sqrt{\frac{1-\cos A}{1+\cos A}}$
$=\sqrt{\frac{1-\cos A}{1+\cos A} \times \frac{1+\cos A}{1+\cos A}}$
$=\sqrt{\frac{1-\cos ^2 A}{(1+\cos A)^2}}$
$=\sqrt{\frac{\sin ^2 A}{(1+\cos A)^2}}$
$=\frac{\sin A}{1+\cos A}$
View full question & answer→Question 852 Marks
If $4 \cos^2 A – 3 = 0$ and $\leq A \leq 90^\circ ,$ then prove that: $\cos 3 A = 4 \cos^3 A – 3 \cos A$
Answer$\text{LHS} = \cos 3A = \cos 90 = 0$
$\text{RHS} = 4 \cos^3 A - 3cos A$
$= 4 \cos^3 30 - 3cos\ 30$
$=4\left(\frac{\sqrt{3}}{2}\right)^3-3\left(\frac{\sqrt{3}}{2}\right)$
$=\frac{3 \sqrt{3}}{2}-\frac{3 \sqrt{3}}{2}=0$
$\text{LHS = LHS}$
View full question & answer→Question 862 Marks
Prove the following identitie:$\sqrt{\frac{1+\sin A}{1-\sin A}}=\frac{\cos A}{1-\sin A}$
Answer$\sqrt{\frac{1+\sin A}{1-\sin A}}$
$=\sqrt{\frac{1+\sin A}{1-\sin A} \times \frac{1-\sin A}{1-\sin A}}$
$=\sqrt{\frac{1-\sin ^2 A}{(1-\sin A)^2}}$
$=\sqrt{\frac{\cos ^2 A}{(1+\sin A)^2}}$
$=\frac{\cos A}{1-\sin A}$
View full question & answer→Question 872 Marks
Prove the following identitie : $\frac{\sin A}{1-\cos A}-\cot A=\operatorname{cosec} A$
Answer$\frac{\sin A}{1-\cos A}-\cot A$
$=\frac{\sin A}{1-\cos A}-\frac{\cos A}{\sin A}$
$=\frac{\sin ^2 A-\cos A+\cos ^2 A}{(1-\cos A) \sin A}$
$=\frac{1}{\sin A}$
$=\operatorname{cosec} A$
View full question & answer→Question 882 Marks
Prove the following identitie:$1-\frac{\sin ^2 A}{1+\cos A}=\cos A$
Answer$1-\frac{\sin ^2 A}{1+\cos A}$
$=\frac{1+\cos A-\sin ^2 A}{1+\cos A}$
$=\frac{\cos A+\cos ^2 A}{1+\cos A}$
$=\frac{\cos A(1+\cos A)}{1+\cos A}$
$=\cos A$
View full question & answer→Question 892 Marks
Prove that $(\sin A + \cos A) (\sec A + \operatorname{cosec A}) = 2 + \sec A \operatorname{cosec A} $
Answer$(\sin A +\cos A )(\sec A +\operatorname{cosec} A )$
$=\frac{\sin A}{\cos A}+1+1+\frac{\cos A}{\sin A}$
$=2+\frac{\cos ^2 A+\sin ^2 A}{\sin A \cos A}$
$=2+\frac{1}{\sin A \cos A}$
$=2+\sec A \operatorname{cosec} A$
View full question & answer→Question 902 Marks
Prove that $(\sin A-\cos A)(1+\tan A+\cot A)=\frac{\sec A}{\operatorname{cosec} 2}-\frac{\operatorname{cosec} A}{\sec ^2 A}$
Answer$(\sin A-\cos A)(1+\tan A+\cot A)$
$=\sin A+\frac{\sin ^2 A}{\cos A}+\cos A-\cos A-\sin A-\frac{\cos ^2 A}{\sin A}$
$=\frac{\sin ^2 A}{\cos A}-\frac{\cos ^2 A}{\sin A}$
$=\frac{\sec A}{\operatorname{cosec} 2}-\frac{\operatorname{cosec} A}{\sec ^2 A}$
View full question & answer→Question 912 Marks
Prove that$\frac{\cot ^2 A}{\operatorname{cosec} A-1}-1=\operatorname{cosec} A$
Answer$\frac{\cot ^2 A}{\operatorname{cosec} A-1}-1$
$=\frac{\cot ^2 A-\operatorname{cosec} A+1}{\operatorname{cosec} A-1}$
$=\frac{-\operatorname{cosec} A+\operatorname{cosec}{ }^2 A}{\operatorname{cosec} A-1}$
$=\frac{\operatorname{cosec} A(\operatorname{cosec} A-1)}{\operatorname{cosec} A-1}$
$=\operatorname{cosec} A$
View full question & answer→Question 922 Marks
Prove that : $\frac{\sin \theta \sin \left(90^{\circ}-\theta\right)}{\cot \left(90^{\circ}-\theta\right)}=1-\sin ^2 \theta$
Answer$\text { LHS }=\frac{\sin \theta \sin \left(90^{\circ}-\theta\right)}{\cot \left(90^{\circ}-\theta\right)}$
$=\frac{\sin \theta \cos \theta}{\tan \theta}$
$=\frac{\sin \theta \cos \theta}{\frac{\sin \theta}{\cos \theta}}$
$\cos ^2 \theta=1-\sin ^2 \theta$
View full question & answer→Question 932 Marks
find the value of angle $A,$ where $0^\circ \leq A \leq 90^\circ .\sin(90^\circ - 3A).\operatorname{cosec} 42^\circ = 1$
Answer$\sin \left(90^{\circ}-3 A\right) \cdot \operatorname{cosec} 42^{\circ}=1$
$\cos 3 A \frac{1}{\sin 42^{\circ}}=1$
$\cos 3 A=\sin 42^{\circ}$
$=\sin \left(90^{\circ}-48^{\circ}\right)$
$=\cos 48^{\circ}$
$3 A=48^{\circ}$
$A=16^{\circ}$
View full question & answer→Question 942 Marks
Find the value of $x,$ if $\sin3x = 2\sin 30^\circ \cos30^\circ$
Answer$\sin 3 x=2 \sin 30^{\circ} \cos 30^{\circ}$
$\sin 3 x=2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$
$\sin 3 x=\frac{\sqrt{3}}{2}=\sin 60^{\circ}$
$3 x=60^{\circ}$
Hence, $x=20^{\circ}$
View full question & answer→Question 952 Marks
Find the value of $x,$ if $\sin \ 2x = 2\sin 45^\circ \cos 45^\circ$
Answer$\sin 2 x=2 \sin 45^{\circ} \cos 45^{\circ}$
$\sin 2 x=2\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)$
$\sin 2x = 1 = \sin 90^\circ$
$2x = 90^\circ$
Hence $, x = 45^\circ$
View full question & answer→Question 962 Marks
Find the value of $x,$ if $\tan x=\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \tan 30^{\circ}}$
Answer$\tan x=\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \tan 30^{\circ}}$
$\tan x=\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3} \cdot \frac{1}{\sqrt{3}}}$
$\tan x=\frac{\frac{3-1}{\sqrt{3}}}{1+1}$
$=\frac{2}{2 \sqrt{3}}=\frac{1}{\sqrt{3}}$
$=\tan 30^{\circ}$
Hence $, x = 30^\circ$
View full question & answer→Question 972 Marks
Find the value of $x,$ if $\cos x = \cos60^\circ \cos30^\circ - \sin60^\circ \sin30^\circ$
Answer$\cos x=\cos 60^{\circ} \cos 30^{\circ}-\sin 60^{\circ} \sin 30^{\circ}$
$\cos x=\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right)$
$\cos x=0=\cos 90^{\circ}$
Hence $, x=90^{\circ}$
View full question & answer→Question 982 Marks
Find the value of $x,$ if $\sin x = \sin60^\circ \cos30^\circ + \cos60^\circ \sin30^\circ$
Answer$\sin x=\sin 60^{\circ} \cos 30^{\circ}+\cos 60^{\circ} \sin 30^{\circ}$
$\sin x=\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$
$\sin x=\frac{3}{4}+\frac{1}{4}=1$
$=\sin 90^{\circ}$
Hence $,x=90^{\circ}$
View full question & answer→Question 992 Marks
Find the value of $x,$ if $\sin x = \sin60^\circ \cos30^\circ - \cos60^\circ \sin30^\circ$
Answer$\sin x=\sin 60^{\circ} \cos 30^{\circ}-\cos 60^{\circ} \sin 30^{\circ}$
$\sin x=\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$
$\sin x=\frac{3}{4}-\frac{1}{4}=\frac{1}{2}$
$=\sin 30^{\circ}$
Hence $,x=30^{\circ}$
View full question & answer→Question 1002 Marks
Evaluate : $\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}}-8 \sin ^2 30^{\circ}$
Answer$\frac{\cos 70^{\circ}}{\sin 20^{\circ}}+\frac{\cos 59^{\circ}}{\sin 31^{\circ}}-8 \sin ^2 30^{\circ}$
$=\frac{\cos \left(90^{\circ}-20^{\circ}\right)}{\sin 20^{\circ}}+\frac{\cos \left(90^{\circ}-31^{\circ}\right)}{\sin 31^{\circ}}-\left(\frac{1}{2}\right)^2$
$=\frac{\sin 20^{\circ}}{\sin 20^{\circ}}+\frac{\sin 31^{\circ}}{\sin 31^{\circ}}-2$
$=1+1-2=0$
View full question & answer→