Question 513 Marks
Prove.
$(\cot A-\operatorname{cosec} A)^2=\frac{1-\cos A}{1+\cos A}$
Answer$\text { R.H.S }=\frac{1-\cos A}{1+\cos A}$
$ =\frac{(1-\cos A)(1-\cos A)}{(1+\cos A)(1-\cos A)}$
v =\frac{(1-\cos A)^2}{1-\cos ^2 A} $
$=\frac{(1-\cos A)^2}{\sin ^2 A} $
$ =\left(\frac{1}{\sin A}-\frac{\cos A}{\sin A}\right)^2$
$=(\operatorname{cosec} A-\cot A)^2$
$=(\cot A-\operatorname{cosec} A)^2 $
$ =\text { L.H.S }$
View full question & answer→Question 523 Marks
Prove.
$\frac{1-\sin A}{1+\sin A}=(\sec A-\tan A)^2$
Answer$\text { LHS }=\frac{1-\sin A}{1+\sin A} $
$ =\frac{1-\sin A}{1+\sin A} \times \frac{1-\sin A}{1-\sin A}$
$ =\frac{(1-\sin A)^2}{1-\sin ^2 A} $
$=\frac{(1-\sin A)^2}{\cos ^2 A} $
$=\left(\frac{1-\sin A}{\cos A}\right)^2 $
$ =\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right)^2$
$ =(\sec A-\tan A)^2$
$=\text { RHS }$
View full question & answer→Question 533 Marks
Prove.
$\frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}=2 \sec A$
Answer$\text { LHS }=\frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}$
$ =\frac{(1+\sin A)^2+\cos ^2 A}{\cos A(1+\sin A)}$
$=\frac{1+\sin ^2 A+2 \sin A+\cos ^2 A}{\cos A(1+\sin A)} $
$=\frac{1+2 \sin A+1}{\cos A(1+\sin A)} $
$=\frac{2(1+\sin A)}{\cos A(1+\sin A)} $
$ =2 \sec A=\text { RHS }$
View full question & answer→Question 543 Marks
Prove.
$\frac{\cot ^2 A}{(\operatorname{cosec} A+1)^2}=\frac{1-\sin A}{1+\sin A}$
Answer$\text { R.H.S }=\frac{1-\sin A}{1+\sin A} $
$ =\frac{1-\frac{1}{\operatorname{cosec} A}}{1+\frac{1}{\operatorname{cosec} A}}=\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1}$
$ =\frac{\operatorname{cosec} A-1}{\operatorname{cosec} A+1} \times \frac{\operatorname{cosec}+1}{\operatorname{cosec} A+1}$
$ =\frac{\operatorname{cosec} A-1}{(\operatorname{cosec} A+1)^2}=\frac{\cot ^2 A}{(\operatorname{cosec} A+1)^2}\left(\because \operatorname{cosec}^2 A-1=\cot ^2 A\right) $
$=\text { L.H.S }$
View full question & answer→Question 553 Marks
Prove.$\frac{1+\cos A}{1-\cos A}=\frac{\tan ^2 A}{(\sec A-1)^2}$
Answer$\text { R.H.S }=\frac{\tan ^2 A }{(\sec A -1)^2}=\frac{\sec ^2 A -1}{(\sec A -1)^2} \quad \ldots\left[\sec ^2 \theta-\tan ^2 \theta=1\right.$
$ \left.\sec ^2 \theta-1=\tan ^2 \theta\right]$
$=\frac{(\sec A +1)(\sec A -1)}{\sec A -1} $
$=\frac{\sec A +1}{\sec A -1}=\frac{\frac{1}{\cos A }+1}{\frac{1}{\cos A }-1}=\frac{\frac{1+\cos A }{\cos A }}{\frac{1-\cos A }{\cos A }} $
$ =\frac{1+\cos A }{1-\cos A } $
$\text { R.H.S. }=\text { L.H.S }$
View full question & answer→Question 563 Marks
Prove.$\frac{\operatorname{cosec} A}{\operatorname{cosec} A-1}+\frac{\operatorname{cosec} A}{\operatorname{cosec} A+1}=2 \sec ^2 A$
Answer$LHS =\frac{\operatorname{cosec} A}{\operatorname{cosec} A-1}+\frac{\operatorname{cosec} A}{\operatorname{cosec} A+1} $
$ =\frac{\operatorname{cosec} A(\operatorname{cosec} A+1)+\operatorname{cosec} A(\operatorname{cosec} A-1)}{(\operatorname{cosec} A-1)(\operatorname{cosec} A+1)} $
$ =\frac{\operatorname{cosec} c^2 A+\operatorname{cosec} A+\operatorname{cosec}{ }^2 A-\operatorname{cosec} A}{(\operatorname{cosec} A)^2-(1)^2} $
$=\frac{2 \operatorname{cosec}{ }^2 A}{\operatorname{cosec} c^2 A-1}$
$ =\frac{2 \operatorname{cosec}{ }^2 A}{\cot ^2 A} \quad \ldots .\left(\because \operatorname{cosec}^2 A -1=\cot ^2 A \right) $
$ =2 \frac{\frac{1}{\sin ^2 A}}{\frac{\cos ^2 A}{\sin ^2 A}} $
$ =\frac{2}{\cos ^2 A}$
$ =2 \sec ^2 A$
$=\text { RHS } $
View full question & answer→Question 573 Marks
prove.
$(sinA + cosecA)^2 + (cosA + secA)^2 = 7 + \tan^2A + \cot^2A$
Answer$LHS =(sinA + cosecA)^2 + (cosA + secA)^2$
$= \sin^2A + cosec^2A + 2 sinA\ cosecA + \cos^2A + sec^2A + 2 cosA secA$
$= \sin^2A + \cos^2A +cosec^2A +sec^2A + 2 + 2$
$= 1 +cosec^2A +sec^2A + 4$
$= (1 + \cot^2A) + (1 + \tan^2A) + 5$
$= 7 + \tan^2A + \cot^2A = RHS$
View full question & answer→Question 583 Marks
Prove.$\frac{\sec A-\tan A}{\sec A+\tan A}=1-2 \sec A \tan A+2 \tan ^2 A$
Answer$\text { LHS }=\frac{\sec A-\tan A}{\sec A+\tan A}$
$=\frac{\sec A-\tan A}{\sec A+\tan A} \times \frac{\sec A-\tan A}{\sec A-\tan A}$
$ =\frac{(\sec A-\tan A)^2}{\sec ^2 A-\tan ^2 A} $
$=\frac{\sec ^2 A+\tan ^2 A-2 \sec A \tan A}{1} $
$ =1+\tan ^2 A+\tan ^2 A-2 \sec A \tan A $
$ =1-2 \sec A \tan A+2 \tan ^2 A=\text { RHS }$
View full question & answer→Question 593 Marks
Prove.$\frac{1}{\sec A+\tan A}=\sec A-\tan A$
Answer$ \text { LHS }=\frac{1}{\sec A +\tan A } $
$ =\frac{1}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}} $
$=\frac{1}{\frac{1+\sin A }{\cos A }}$
$ =\frac{\cos A }{1+\sin A } \times \frac{1-\sin A }{1+\sin A } $
$=\frac{\cos A(1-\sin A)}{(1)^2-\sin 2 A}$
$ =\frac{\cos A(1-\sin A)}{\cos ^2 A} $
$ =\frac{1}{\cos A}-\frac{\sin A}{\cos A} $
$ =\sec A -\tan A $
LHS $=$ RHS
Hence proved.
View full question & answer→Question 603 Marks
Prove.
$(\operatorname{cosec} A-\sin A )(\sec A -\cos A )(\tan A +\cot A )= 1$
Answer$\text { LHS }=(\operatorname{cosec} A-\sin A )(\sec A -\cos A )(\tan A +\cot A ) $
$=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)\left(\frac{1}{\tan A}+\tan A\right) $
$ =\left(\frac{1-\sin ^2 A}{\sin A}\right)\left(\frac{1-\cos ^2 A}{\cos A}\right)\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)$
$ =\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right)\left(\frac{\sin ^2 A+\cos ^2 A}{\sin A \cdot \cos A}\right) $
$ =\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right)\left(\frac{1}{\sin A \cdot \cos A}\right) $
$=1=\text { RHS }$
View full question & answer→Question 613 Marks
If $A$ and $B$ are complementary angles, prove that: $cosec^2A + cosec^2B = cosec^2A \ cosec^2B$
AnswerSince, $A$ and $B$ are complementary angles, $A + B = 90^\circ$
$cosec^2A + cosec^2B$
$= cosec^2A + (cosec(90^\circ - A))^2$
$= cosec^2A + sec^2A$
$=\frac{1}{\sin ^2 A}+\frac{1}{\cos ^2 A}$
$=\frac{\cos ^2 A+\sin ^2 A}{\sin ^2 A \cos ^2 A}$
$=\frac{1}{\sin ^2 A \cos ^2 A}$
$= cosec^2A \ (sec(90A^\circ - B))^2$
$= cosec^2A \ cosec^2B$
View full question & answer→Question 623 Marks
If $A$ and $B$ are complementary angles, prove that: $\cot B + \cos B = \sec A \cos B (1 + \sin B)$
AnswerSince, $A$ and $B$ are complementary angles, $A + B = 90^\circ$
$=\cot B + \cos B$
$= \cot(90^\circ - A) + \cos(90^\circ - A)$
$= \tan A + \sin A$
$=\frac{\sin A}{\cos A}+\sin A$
$=\frac{\sin A+\sin A \cos A}{\cos A}$
$=\frac{\sin A(1+\cos A)}{\cos A}$
$= \sec A \sin A (1 + \cos A)$
$= \sec A \sin (90^\circ - B)(1 + \cos(90^\circ - B))$
$= \sec A \cos B(1 + \sin B)$
View full question & answer→Question 633 Marks
Prove that: $\frac{1}{1+\sin \left(90^{\circ}-A\right)}+\frac{1}{1-\sin \left(90^{\circ}-A\right)}=2 \sec ^2\left(90^{\circ}-A\right)$
Answer$\frac{1}{1+\sin \left(90^{\circ}-A\right)}+\frac{1}{1-\sin \left(90^{\circ}-A\right)}$
$=\frac{1}{1+\cos A}+\frac{1}{1-\cos A}$
$=\frac{1-\cos A+1+\cos A}{(1+\cos A)(1-\cos A)}$
$=\frac{2}{1-\cos ^2 A}$
$= 2cosec^2A$
$= 2 \ sec^2(90^\circ - A)$
View full question & answer→Question 643 Marks
Prove that:$\frac{1}{1+\cos \left(90^{\circ}-A\right)}+\frac{1}{1-\cos \left(90^{\circ}-A\right)}=2 \operatorname{cosec} 2\left(90^{\circ}-A\right)$
Answer$\begin{aligned} & \frac{1}{1+\cos \left(90^{\circ}-A\right)}+\frac{1}{1-\cos \left(90^{\circ}-A\right)} \\ & =\frac{1}{1+\sin A}+\frac{1}{1-\sin A} \\ & \frac{1-\sin A+1+\sin A}{(1+\sin A)(1-\sin A)} \\ & =\frac{2}{1-\sin ^2 A} \\ & =\frac{2}{\cos ^2 A} \\ & =2 \sec ^2 A \\ & =2 \operatorname{cosec}^2\left(90^{\circ}- A \right)\end{aligned}$
View full question & answer→Question 653 Marks
Evaluate $2\left(\frac{\tan 35^{\circ}}{\cot 55^{\circ}}\right)+\left(\frac{\cot 55^{\circ}}{\tan 35^{\circ}}\right)-3\left(\frac{\sec 40^{\circ}}{\operatorname{cosec} 50^{\circ}}\right)$
Answer$=2\left(\frac{\tan 35^{\circ}}{\cot 55^{\circ}}\right)+\left(\frac{\cot 55^{\circ}}{\tan 35^{\circ}}\right)-3\left(\frac{\sec 40^{\circ}}{\operatorname{cosec} 50^{\circ}}\right)$
$=2\left(\frac{\tan \left(90^{\circ}-55^{\circ}\right)}{\cot 55^{\circ}}\right)+\left(\frac{\cot \left(90^{\circ}-35^{\circ}\right)}{\tan 35^{\circ}}\right)-3\left(\frac{\sec \left(90^{\circ}-50^{\circ}\right)}{\operatorname{cosec} 50^{\circ}}\right)$
$=2\left(\frac{\cot 55^{\circ}}{\cot 55^{\circ}}\right)+\left(\frac{\tan 35^{\circ}}{\tan 35^{\circ}}\right)-3\left(\frac{\cos e c 50^{\circ}}{\operatorname{cosec} 50^{\circ}}\right)$
$=2(1)^2+1^2+-3$
$=2+1-3$
$=0$
View full question & answer→Question 663 Marks
If $4 \ cos2 \ A – 3 = 0,$ Show that: $\cos 3 \ A = 4 \cos^3 \ A – 3 \cos A$
Answer$4cos^2A - 3 = 0$
$\Rightarrow 4 \cos^2A = 3$
$\Rightarrow \cos^2A = 3/4$
$\Rightarrow \cos A=\frac{\sqrt{3}}{2}$
so, $A = 30^\circ$
$ \text{LHS} = \cos^3A = \cos 90^\circ = 0$
$ \text{RHS} = 4 \cos^3A - 3 \ cosA$
$= 4cos^330^\circ - 3 \ cos30^\circ$
$=4\left(\frac{\sqrt{3}}{2}\right)^3-3\left(\frac{\sqrt{3}}{2}\right)$
$=\frac{3 \sqrt{3}}{2}-\frac{3 \sqrt{3}}{2}=0$
$\text { LHS }=\text { RHS }$
View full question & answer→Question 673 Marks
If $2 \sin A – 1 = 0,$ show that: $Sin \ 3A = 3 \sin A – 4 \sin^3A$
Answer$2 \ sinA − 1 = 0$
$\Rightarrow sinA = 1/2$
We know $sin30^\circ = 1/2$
So, $A = 30^\circ$
$ \text{LHS} = \sin 3A = sin90^\circ = 1$
$ \text{RHS} = 3 sinA - 4sin^3A$
$=3 sin30^\circ - 4sin^330^\circ$
$=3\left(\frac{1}{2}\right)-4\left(\frac{1}{2}\right)^3$
$=\frac{3}{2}-\frac{1}{2}=1$
$\text { LHS }=\text { RHS }$
View full question & answer→Question 683 Marks
If $\sec A+\tan A=p$, show that: $\sin A =\frac{p^2-1}{p^2+1}$
Answer$\frac{p^2-1}{p^2+1}$
$=\frac{(\sec A +\tan A )^2-1}{(\sec A +\tan A )^2+1}$
$=\frac{\sec ^2 A +\tan ^2 A +2 \tan A \sec A -1}{\sec ^2 A +\tan ^2 A +2 \tan A \sec A +1}$
$=\frac{\tan ^2 A +\tan ^2 A +2 \tan A \sec A }{\sec ^2 A +\sec ^2 A +2 \tan A \sec A }$
$=\frac{2 \tan ^2 A +2 \tan A \sec A }{2 \sec ^2 A +2 \tan A \sec A }$
$=\frac{2 \tan A (\tan A +\sec A )}{2 \sec A (\tan A +\sec A )}$
$=\sin A $
View full question & answer→Question 693 Marks
Evaluate without using trigonometric tables,$\sin ^2 28^{\circ}+\sin ^2 62^{\circ}+\tan ^2 38^{\circ}-\cot ^2 52^{\circ}+\frac{1}{4} \sec ^2 30^{\circ}$
Answer$\sin ^2 28^{\circ}+\sin ^2 62^{\circ}+\tan ^2 38^{\circ}-\cot ^2 52^{\circ}+\frac{1}{4} \sec ^2 30^{\circ}$
$=\sin ^2 28^{\circ}+\sin ^2\left(90^{\circ}-28^{\circ}\right)+\tan ^2 38^{\circ}-\cot ^2\left(90^{\circ}-38^{\circ}\right)+\frac{1}{4} \sec ^2 30^{\circ}$
$=\left(\sin ^2 28^{\circ}+\cos ^2 28^{\circ}\right)+\tan ^2 38^{\circ}-\tan ^2 38^{\circ}+\frac{1}{4} \times\left(\frac{2}{\sqrt{3}}\right)^2$
$=1+0+\frac{1}{4} \times \frac{4}{3}$
$=1+\frac{1}{3}$
$=\frac{4}{3}$
View full question & answer→Question 703 Marks
Prove the identity $(\sin \theta + \cos \theta ) (\tan \theta + \cot \theta ) = \sec \theta + \operatorname{cosec} \theta .$
Answer$\text { L.H.S. }=(\sin \theta+\cos \theta)(\tan \theta+\cot \theta)$
$=(\sin \theta+\cos \theta)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)$
$=(\sin \theta+\cos \theta)\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \sin \theta}\right)$
$=(\sin \theta+\cos \theta) \times \frac{1}{\sin \theta \cos \theta}\left(\because \sin ^2 \theta+\cos ^2 \theta=1\right)$
$=\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta}$
$=\frac{\sin \theta}{\cos \theta \sin \theta}+\frac{\cos \theta}{\cos \theta \sin \theta}$
$=\frac{1}{\cos \theta}+\frac{1}{\sin \theta}$
$=\sec \theta+\operatorname{cosec} \theta$
$= \text{R.H.S}$
Hence proved.
View full question & answer→Question 713 Marks
Prove that: $(cosecA - sinA) (secA - cosA) sec^2A = tanA$
Answer$ \text{LHS} \ (cosecA - sinA) (secA - cosA) sec^2A$
$=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right) \sec ^2 A$
$=\left(\frac{1-\sin ^2 A}{\sin A}\right)\left(\frac{1-\cos ^2 A}{\cos A}\right) \sec ^2 A$
$=\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right) \sec ^2 A$
$=\frac{\sin A}{\cos A}=\tan A= \text{RHS}$
View full question & answer→Question 723 Marks
If $0^\circ < A < 90^\circ ;$ Find $A,$ if $\frac{\sin A}{\sec A-1}+\frac{\sin A}{\sec A+1}=2$
Answer$\frac{\sin A}{\sec A-1}+\frac{\sin A}{\sec A+1}=2$
$\Rightarrow \frac{\sin A \sec A+\sin A+\sec A \sin A-\sin A}{(\sec A-1)(\sec A+1)}=2$
$\Rightarrow \frac{2 \sin A \sec A}{\sec ^2 A-1}=2$
$\Rightarrow \frac{\sin A \sec A}{\tan ^2 A}=1$
$\Rightarrow \frac{\cos A}{\sin A}=1$
$\Rightarrow \cot A=1$
We know $\cot 45^\circ = 1$
Hence, $A = 45^\circ$
View full question & answer→Question 733 Marks
If $0^\circ < A < 90^\circ ;$ Find $A,$ if :$\frac{\cos A}{1-\sin A}+\frac{\cos A}{1+\sin A}=4$
Answer$\frac{\cos A}{1-\sin A}+\frac{\cos A}{1+\sin A}=4$
$\Rightarrow \frac{\cos A+\cos A \sin A+\cos A-\sin A \cos A}{(1-\sin A)(1+\sin A)}=4$
$\Rightarrow \frac{2 \cos A}{1-\sin A}=4$
$\Rightarrow \frac{2 \cos A}{\cos ^2 A}=4$
$\Rightarrow \frac{1}{\cos A}=2$
$\Rightarrow \cos A=\frac{1}{2}$
We know $\cos 60=\frac{1}{2}$
Hence, $A=60$
View full question & answer→Question 743 Marks
Find $A,$ if $0^\circ \leq A \leq 90^\circ$ and $2 \cos^2 A+ \cos A - 1 = 0$
Answer$2 \cos ^2A+ \cos A - 1 = 0$
$\Rightarrow 2 \cos ^2A + 2 \cos A - \cos A - 1 = 0$
$\Rightarrow 2 \cos A ( \cos A + 1) - 1( \cos A + 1) = 0$
$\Rightarrow (2 \cos A - 1)( \cos a + 1) = 0$
$\Rightarrow \cos A = 1/2$ or $ \cos A = -1$
We know $\cos 60^\circ = 1/2$
We also know that for no value of $A(0^\circ \leq A\leq 90^\circ ), \cos A = -1.$
Hence, $A = 60^\circ$
View full question & answer→Question 753 Marks
Prove the following identitie:
$(1 + \tan A + \sec A)(1 + \cot A - \cos ec A) = 2$
Answer$(1 + \tan A + \sec A) (1 + \cot A - \cos ec A)$
$=1 + \cot A - cosec A + \tan A + 1 - \sec A + \sec A + cosec A - cosec A \sec A$
$=2+\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}-\frac{1}{\sin A \cos A}$
$=2+\frac{\cos ^2 A+\sin ^2 A}{\sin A \cos A}-\frac{1}{\sin A \cos A}$
$=2+\frac{1}{\sin A \cos A}-\frac{1}{\sin A \cos A}$
$=2$
View full question & answer→Question 763 Marks
Prove the following identitie: $cosec^4A (1 - \cos^4A) - 2 \cot^2A = 1$
Answer$cosec^4A (1 - \cos^4A) - 2 \cot^2A$
$=cosec^4A (1 - \cos^2A) (1 + \cos^2A) - 2 \cot^2A$
$= cosec^4A (\sin^2A) (1 + \cos^2A) - 2 \cot^2A$
$= cosec^2A (1 + \cos^2A) - 2 \cot^2A$
$=\cos e c^2 A+\frac{\cos ^2 A}{\sin ^2 A}-2 \cot ^2 A$
$= cosec^2A + \cot^2A - 2cot^2A$
$= cosec^2A - \cot^2A$
$= 1$
View full question & answer→Question 773 Marks
Prove the following identitie : $\sec^4A (1 - \sin^4A) - 2 \tan^2A = 1$
Answer$ \sec ^4 A\left(1-\sin ^4 A\right)-2 \tan ^2 A$
$ =\sec ^4 A-\sec ^4 A \sin ^4 A-2 \tan ^2 A$
$ =\left(\sec ^2 A\right)^2-\frac{1}{\cos ^4 A} \sin ^4 A-2 \tan ^2 A$
$ =(1+\tan ^2 A)^2-\tan ^4 A-2 \tan ^2 A \ldots[\sec ^2 A-\tan ^2 A=1\sec ^2 A=1+\tan ^2 A]$
$=(1)^2+\left(\tan ^2 A\right)^2-2 \times 1 \times \tan ^2 A-\tan ^4 A-2 \tan ^2 A$
$ =1+\tan ^4 A+2 \tan ^2 A-\tan ^4 A-2 \tan ^2 A$
$ =1$
View full question & answer→Question 783 Marks
Prove the following identitie:$\frac{\left(1-2 \sin ^2 A\right)^2}{\cos ^4 A-\sin ^4 A}=2 \cos ^2 A-1$
Answer$\frac{\left(1-2 \sin ^2 A\right)^2}{\cos ^4 A-\sin ^4 A}$
$=\frac{\left(1-2 \sin ^2 A\right)^2}{\left(\cos ^2 A-\sin ^2 A\right)\left(\cos ^2 A+\sin ^2 A\right)}$
$=\frac{\left(1-2 \sin ^2 A\right)^2}{1-\sin ^2 A-\sin ^2 A}$
$=\frac{\left(1-2 \sin ^2 A\right)^2}{1-2 \sin ^2 A}$
$=1-2 \sin ^2 A$
$=1-2\left(1-\cos ^2 A\right)$
$=2 \cos ^2 A-1$
View full question & answer→Question 793 Marks
If $4 \cos^2 A – 3 = 0$ and $\leq A \leq 90^\circ ,$ then prove that : $\sin 3 A = 3 \sin A – 4 \sin^3 A$
Answer$4 \cos^2A − 3 = 0$
$\cos A=\frac{\sqrt{ } 3}{2}$
We know $\cos 30=\frac{\sqrt{3}}{2}$
So, $A = 30$
$(i) \ \text{LHS} = \sin 3A = \sin 90 = 1$
$ \text{RHS} = 3sin A - 4 \sin^3 A$
$= 3sin 30 - 4 \sin^3 30$
$=3 \times \frac{1}{2}-4 \times\left(\frac{1}{2}\right)^3$
$=\frac{3}{2}-\frac{1}{2}$
$=1$
$\text { LHS }=\text { RHS }$
View full question & answer→Question 803 Marks
Prove the following identitie:$\frac{1+(\sec A-\tan A)^2}{\operatorname{cosec} A(\sec A-\tan A)}=2 \tan A$
Answer$\frac{1+(\sec A-\tan A)^2}{\operatorname{cosec} A(\sec A-\tan A)}$
$=\frac{\left(\sec ^2 A-\tan ^2 A\right)+(\sec A-\tan A)^2}{\operatorname{cosec} A(\sec A-\tan A)}$
$=\frac{(\sec A-\tan A)(\sec A+\tan A)+(\sec A+\tan A)^2}{\operatorname{cosec} A(\sec A-\tan A)}$
$=\frac{(\sec A-\tan A)+(\sec A-\tan A)}{\operatorname{cosec} A}$
$=\frac{2 \sec A}{\operatorname{cosec} A}$
$=2 \frac{\frac{1}{\cos A}}{\frac{1}{\sin A}}$
$=2 \tan A$
View full question & answer→Question 813 Marks
Prove the following identitie:$\frac{\cos A}{1+\sin A}+\tan A=\sec A$
Answer$\frac{\cos A}{1+\sin A}+\tan A$
$=\frac{\cos A}{1+\sin A}+\frac{\sin A}{\cos A}$
$=\frac{\cos ^2 A+\sin A+\sin ^2 A}{(1+\sin ) \cos A}$
$=\frac{1+\sin A}{(1+\sin A) \cos A}$
$=\frac{\cos ^3 A+\cos A \sin A-\sin ^2 A}{\cos ^2 A-\sin A \cos A}$
$=\frac{1}{\cos A}$
$=\sec A$
View full question & answer→Question 823 Marks
Prove the following identitie:$\frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A}=1+\tan A+\cot A$
Answer$\frac{\cot A}{1-\tan A}+\frac{\tan A}{1-\cot A}$
$=\frac{\frac{1}{\tan A}}{1-\tan A}+\frac{\tan A}{1-\frac{1}{\tan A}}$
$=\frac{1}{\tan A(1-\tan A)}+\frac{\tan ^2 A}{\tan A-1}$
$=\frac{1-\tan ^3 A}{\tan A(1-\tan A)}$
$=\frac{(1-\tan A)\left(1+\tan A+\tan ^2 A\right)}{\tan A(1-\tan A)}$
$=\frac{1+\tan A+\tan ^2 A}{\tan A}$
$=\cot A +1+\tan A $
View full question & answer→Question 833 Marks
Prove the following identitie:$\frac{1-\cos A}{\sin A}+\frac{\sin A}{1-\cos A}=2 \operatorname{cosec} A$
Answer$\frac{1-\cos A}{\sin A}+\frac{\sin A}{1-\cos A}$
$=\frac{(1-\cos A)^2+\sin ^2 A}{\sin A(1-\cos A)}$
$=\frac{1+\cos ^2 A-2 \cos A+\sin ^2 A}{\sin A(1-\cos A)}$
$\frac{2-2 \cos A}{\sin A(1-\cos A)}$
$\frac{2(1-\cos A)}{\sin A(1-\cos A)}$
$=2 \operatorname{cosec} A$
View full question & answer→Question 843 Marks
Prove the following identitie:$\operatorname{cosec} A-\cot A=\frac{\sin A}{1+\cos A}$
Answer$\operatorname{cosec} A-\cot A$
$=\frac{1}{\sin A}-\frac{\cos A}{\sin A}$
$=\frac{1-\cos A}{\sin A}$
$=\frac{1-\cos A}{\sin A} \times \frac{1+\cos A}{1+\cos A}$
$=\frac{1-\cos ^2 A}{\sin A(1+\cos A)}$
$\frac{\sin 2}{\sin A(1+\cos A)}$
$=\frac{\sin A}{1+\cos A}$
View full question & answer→Question 853 Marks
Prove that$\frac{\cot A-1}{2-\sec ^2 A}=\frac{\cot A}{1+\tan A}$
Answer$\text { L.H.S. }=\frac{\cot A -1}{2-\sec ^2 A }$
$=\frac{\frac{1}{\tan A }-1}{2-\left(1+\tan ^2 A \right)}$
$=\frac{1-\tan A }{\tan A \left(1-\tan ^2 A \right)}$
$=\frac{1-\tan A }{\tan A (1-\tan A )(1+\tan A )}$
$=\frac{1}{\tan A (1+\tan A )}$
$=\frac{1}{\tan A } \times \frac{1}{1+\tan A }$
$=\frac{\cot A }{1+\tan A }$
$\text{R.H.S.}$
Hence proved.
View full question & answer→Question 863 Marks
Prove the following identitie:$\frac{1}{\cos A+\sin A}+\frac{1}{\cos A-\sin A}=\frac{2 \cos A}{2 \cos ^2 A-1}$
Answer$\frac{1}{\cos A+\sin A}+\frac{1}{\cos A-\sin A}$
$=\frac{\cos A+\sin A+\cos A-\sin A}{(\cos A+\sin A)(\cos A-\sin A)}$
$=\frac{2 \cos A}{\cos ^2 A-\sin ^2 A}$
$=\frac{2 \cos A}{\cos ^2 A-\left(1-\cos ^2 A\right)}$
$=\frac{2 \cos A}{2 \cos ^2 A-1}$
View full question & answer→Question 873 Marks
Prove that
$(\tan A + \cot A ) (\operatorname{cosec} A - \sin A ) (sec A - \cos A ) = 1$
Answer$(\tan A+\cot A)(\operatorname{cosec} A-\sin A)(\sec A-\cos A)$
$=\left(\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)$
$=\left(\frac{\sin ^2 A+\cos ^2 A}{\sin A \cos A}\right)\left(\frac{1-\sin ^2 A}{\sin A}\right)\left(\frac{1-\cos ^2 A}{\cos A}\right)$
$=\left(\frac{1}{\sin A \cos A}\right)\left(\frac{\cos ^2 A}{\sin A}\right)\left(\frac{\sin ^2 A}{\cos A}\right)$
$=1$
View full question & answer→Question 883 Marks
Prove that
$\cos A (1 + \cot A) + \sin A (1 + \tan A = \sec A + \operatorname{cosec} A)$
Answer$\cos A (1+\cot A )+\sin A (1+\tan A )$
$=\cos A+\frac{\cos ^2 A}{\sin A}+\sin A+\frac{\sin ^2 A}{\cos A}$
$=\sin A+\frac{\cos ^2 A}{\sin A}+\cos A+\frac{\sin ^2 A}{\cos A}$
$=\left(\frac{\cos ^2 A+\sin ^2 A}{\sin A}\right)+\left(\frac{\cos ^2 A+\sin ^2 A}{\cos A}\right)$
$=\frac{1}{\sin A}+\frac{1}{\cos A}$
$=\operatorname{cosec} A+\sec A$
View full question & answer→Question 893 Marks
Prove that$\frac{\cos A}{1+\sin A}=\sec A-\tan A$
Answer$\begin{aligned} & \frac{\cos A}{1+\sin A} \\ = & \frac{\cos A}{1+\sin A} \times \frac{1-\sin A}{1-\sin A} \\ = & \frac{\cos A(1-\sin A)}{1-\sin ^2 A} \\ = & \frac{\cos A(1-\sin A)}{\cos ^2 A} \\ = & \frac{1-\sin A}{\cos A} \\ = & \sec A-\tan A\end{aligned}$
View full question & answer→Question 903 Marks
Prove that$\frac{1}{\sin A-\cos A}-\frac{1}{\sin A+\cos A}=\frac{2 \cos A}{2 \sin ^2 A-1}$
Answer$\frac{1}{\sin A-\cos A}-\frac{1}{\sin A+\cos A}$
$=\frac{\sin A+\cos A-\sin A+\cos A}{(\sin A-\cos A)(\sin A+\cos A)}$
$=\frac{2 \cos A}{\sin ^2 A-\cos ^2 A}$
$=\frac{2 \cos A}{\sin ^2 A-\left(1-\sin ^2 A\right)}$
$=\frac{2 \cos A}{2 \sin ^2 A-1}$
View full question & answer→Question 913 Marks
Find the value of x, if $\cos(2x - 6) = \cos^230^\circ - \cos^260^\circ$
Answer$\cos(2x - 6) = \cos^230^\circ - \cos^260^\circ$
$\cos(2x - 6) = \cos^2 (90^\circ - 60^\circ ) - \cos^2 60^\circ$
$\cos(2x - 6) = \sin^2 60^\circ - \cos^2 60^\circ$
$\cos (2 x-6)=1-2 \cos ^2 60^{\circ}=1-2\left(\frac{1}{2}\right)^2=1-\frac{1}{2}=\frac{1}{2}$
$\cos(2x - 6) = 1/2$
$\cos(2x - 6) = cos\ 60^\circ$
$(2x - 6) = 60^\circ$
$2x = 66^\circ$
Hence, $x = 33^\circ$
View full question & answer→Question 923 Marks
A triangle $\text{ABC}$ is right angles at $B;$ find the value of
$\frac{\sec A \operatorname{cosec} A-\tan A \cot C}{\sin B}$
AnswerSince, $\text{ABC}$ is a right angled triangle, right angled at $B.$
so, $A + C = 90^\circ$
$\frac{\sec A \operatorname{cosec} A-\tan A \cot C}{\sin B}$
$=\frac{\sec \left(90^{\circ}-C\right) \cdot \cos e c C-\tan \left(90^{\circ}-C\right) \cdot \cot C}{\sin 90^{\circ}}$
$=\frac{\cos e c C \cdot \cos e c C-\cot C \cdot \cot C}{1}$
$=1\left(\because \operatorname{cosec}^2 \theta-\cot ^2 \theta=1\right)$
View full question & answer→Question 933 Marks
Show that: $\sin A \cos A-\frac{\sin A \cos \left(90^{\circ}-A\right) \cos A}{\sec \left(90^{\circ}-A\right)}-\frac{\cos A \sin \left(90^{\circ}-A\right) \sin A}{\operatorname{cosec}\left(90^{\circ}-A\right)}=0$
Answer$\sin A \cos A-\frac{\sin A \cos \left(90^{\circ}-A\right) \cos A}{\sec \left(90^{\circ}-A\right)}-\frac{\cos A \sin \left(90^{\circ}-A\right) \sin A}{\operatorname{cosec}\left(90^{\circ}-A\right)}$
$=\sin A \cos A-\frac{\sin A \sin A \cos A}{\cos e c A}-\frac{\cos A \cos A \sin A}{\sec A}$
$= sinA \ cosA - \sin^3A \ cosA - \cos^3A \ sinA$
$= sinA \ cosA - sinA \ cosA (\sin^2A + \cos^2A)$
$= sinA \ cosA - sinA \ cosA(1)$
$= 0$
View full question & answer→Question 943 Marks
Without using trigonometrical tables, evaluate:$\operatorname{cosec} 257^{\circ}-\tan ^2 33^{\circ}+\cos 44^{\circ} \operatorname{cosec} 46^{\circ}-\sqrt{2} \cos 45^{\circ}-\tan ^2 60^{\circ}$
Answer$\operatorname{cosec}^2 57^{\circ}-\tan ^2 33^{\circ}+\cos 44^{\circ} \operatorname{cosec} 46^{\circ}-\sqrt{2} \cos 45^{\circ}-\tan ^2 60^{\circ}$
$=\operatorname{cosec}^2\left(90^{\circ}-33^{\circ}\right)^{\circ}-\tan ^2 33^{\circ}+\cos 44^{\circ} \operatorname{cosec}\left(90^{\circ}-44^{\circ}\right)-\sqrt{2} \cos 45^{\circ}-\tan ^2 60^{\circ}$
$=\sec ^2 33^{\circ}-\tan ^2 33^{\circ}+\cos 44^{\circ} \sec 44^{\circ}-\sqrt{2} \cos 45^{\circ}-\tan ^2 60$
$=1+1-\sqrt{2} \cos 45^{\circ}-\tan ^2 60$
$=1+1-\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)-(\sqrt{3})^2$
$=2-1-3$
$=-2$
View full question & answer→Question 953 Marks
Evaluate $\sin^234^\circ + \sin^256^\circ + 2tan \ 18^\circ \ tan72^\circ - \cot^230^\circ$
Answer$\sin ^2 34^{\circ}+\sin ^2 56^{\circ}+2 \tan 18^{\circ} \tan 72^{\circ}-\cot ^2 30^{\circ}$
$=\sin ^2 34^{\circ}+\sin ^2\left(90^{\circ}-34^{\circ}\right)+23 \tan 18^{\circ} \tan \left(90^{\circ}-72^{\circ}\right)-\cot ^2 30^{\circ}$
$=\sin ^2 34^{\circ}+\cos ^2 34^{\circ}+2 \tan 18^{\circ} \cot ^2 18^{\circ}-\cot ^2 30^{\circ}$
$=\left(\sin ^2 34^{\circ}+\cos ^2 34^{\circ}\right)+2 \tan 18^{\circ} \times \frac{1}{\tan 18^{\circ}}-\cot ^2 30^{\circ}$
$=1+2 \times 1-(\sqrt{3})^2$
$=1+2-3$
$=3-3$
$=0$
View full question & answer→Question 963 Marks
Evaluate:$\frac{\sin 35^{\circ} \cos 55^{\circ}+\cos 35^{\circ} \sin 55^{\circ}}{\cos e c^2 10^{\circ}-\tan ^2 80^{\circ}}$
Answer$\frac{\sin 35^{\circ} \cos 55^{\circ}+\cos 35^{\circ} \sin 55^{\circ}}{\cos e c^2 10^{\circ}-\tan ^2 80^{\circ}}$
$\frac{\sin 35^{\circ} \cdot \cos \left(90^{\circ}-35^{\circ}\right)+\cos 35^{\circ} \cdot \sin \left(90^{\circ}-35^{\circ}\right)}{\cos e c^2\left(90^{\circ}-80^{\circ}\right)-\tan ^2 80^{\circ}}$
$=\frac{\sin 35^{\circ} \cdot \sin 35^{\circ}+\cos 35^{\circ} \cdot \cos 35^{\circ}}{\sec ^2 80^{\circ}-\tan ^2 80^{\circ}}$
$=\frac{\sin ^2 35^{\circ}+\cos ^2 35^{\circ}}{\sec ^2 80^{\circ}-\tan ^2 80^{\circ}}=\frac{1}{1}=1$
View full question & answer→Question 973 Marks
If $\frac{\cos A}{\cos B}=m$ and $\frac{\cos A}{\sin B}= n$ show that: $\left(m^2+n^2\right) \cos ^2 B=n^2$.
Answer$\text { LHS }=\left(m^2+n^2\right) \cos ^2 B$
$=\left(\frac{\cos ^2 A}{\cos ^2 B}+\frac{\cos ^2 A}{\sin ^2 B}\right) \cos ^2 B$
$=\left(\frac{\cos ^2 A \sin ^2 B+\cos ^2 A \cos ^2 B}{\cos ^2 B \sin ^2 B}\right) \cos ^2 B$
$=\left(\frac{\cos ^2 A \sin ^2 B+\cos ^2 A \cos ^2 B}{\sin ^2 B}\right)$
$=\frac{\cos ^2 A\left(\sin ^2 B+\cos ^2 B\right)}{\sin ^2 B}$
$=\frac{\cos ^2 A}{\sin ^2 B}$
$=n^2$
Hence, $(m^2 + n^2) \cos^2B = n^2.$
View full question & answer→Question 983 Marks
If $\sin A + \cos A = m$ and $ \sec A + cosec A = n,$ show that: $n (m^2 - 1) = 2m$
Answer$\sin A + \cos A = m$ and $ \sec A + co sec A = n$
Consider $ \text{L.H.S}$
$=n (m2 - 1)$
$=(\sec A+\cos e c A)\left((\sin A+\cos A)^2-1\right)$
$=\left(\frac{1}{\cos A}+\frac{1}{\sin A}\right)\left(\sin ^2 A+\cos ^2 A+2 \sin A \cos A-1\right)$
$=\left(\frac{\cos A+\sin A}{\sin A \cos A}\right)(1+2 \sin A \cos A-1)$
$=\frac{\cos A+\sin A}{\sin A \cos A}(2 \sin A \cos A)$
$= 2 ( \sin A + \cos A)$
$= 2m = \text{RHS}$
View full question & answer→Question 993 Marks
If $m = a \ sec A + b \ \tan A$ and $n = a \ \tan A + b \ sec A,$ then prove that $: m^2 - n^2 = a^2 - b^2$
AnswerGiven,
$m = a \ secA + b \ tanA$ and $n = a \ tanA + b \ secA$
$m^2 - n^2= (a \ secA + b \ tanA)^2 - ( a \ tanA + b \ secA)^2$
$= a^2 \ sec^2A + b^2tan^2A + 2ab \ secA \ tanA - (a^2 \tan^2A + b^2 \ sec^2A + 2ab \ secA \ tanA)$
$= sec^2A (a^2 - b^2) + \tan^2A (b^2 - a^2)$
$= (a^2 - b^2) [sec^2A - \tan^2A]$
$=(a^2 - b^2) [ sincesec^2A - \tan^2A = 1]$
Hence, $m^2 - n^2 = a^2 - b^2$
View full question & answer→Question 1003 Marks
If $x \cos A + \sin A = m$ and $X \sin A – y \cos A = n,$ then prove that: $x^2 + y^2 = m^2 + n^2$
Answer$m^2 + n^2$
$= (x \ cosA + y \ sinA)^2 + (x \ sinA - y \ cosA)^2$
$= x^2cos^2A+y^2sin^2A+2xy \ sinA \ cosA+x^2 \sin^2A+y^2cos^2A - 2xy \ sinA \ cosA$
$= x^2(\cos^2A + \sin^2A) + y^2(\cos^2A + \sin^2A)$
$= x^2 + y^2$
Hence, $x^2 + y^2 = m^2 + n^2$
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