Question 12 Marks
Prove that $\sqrt{2+\tan ^2 \theta+\cot ^2 \theta}=\tan \theta+\cot \theta$.
Answer$\text { LHS }=\sqrt{2+\tan ^2 \theta+\cot ^2 \theta}$
$=\sqrt{\tan ^2 \theta+\cot ^2 \theta+2 \tan \theta \cdot \cot \theta} \quad \ldots[\because \tan \theta \cdot \cot \theta=1]$
$=\sqrt{\tan ^2 \theta+\cot ^2 \theta}$
$= \tan \theta + \cot \theta$
$= \text{RHS}$
Hence proved.
View full question & answer→Question 22 Marks
Prove that:
$\frac{\cos ^3 \theta+\sin ^3 \theta}{\cos \theta+\sin \theta}+\frac{\cos ^3 \theta-\sin ^3 \theta}{\cos \theta-\sin \theta}=2$
Answer$\text { LHS }=\frac{\cos ^3 \theta+\sin ^3 \theta}{\cos \theta+\sin \theta}+\frac{\cos ^3 \theta-\sin ^3 \theta}{\cos \theta-\sin \theta}$
$=\frac{(\cos \theta+\sin \theta)\left(\cos ^2 \theta+\sin ^2 \theta-\cos \theta \sin \theta\right)}{\cos \theta+\sin \theta}+\frac{(\cos \theta-\sin \theta)\left(\cos ^2 \theta+\sin ^2 \theta-\cos \theta \sin \theta\right)}{\cos \theta-\sin \theta}$
$= 1 - \sin \theta \cos \theta + 1 + \sin \theta \cos \theta $
$= 2$
$= RHS$
Hence proved.
View full question & answer→Question 32 Marks
Prove the following identity : $\frac{\left(1+\tan ^2 A\right) \cot A}{\cos e c^2 A}=\tan A$
Answer$\frac{\left(1+\tan ^2 A\right) \cot A}{\cos e c^2 A}$
$=\frac{\sec ^2 A \cot A}{\operatorname{cosec} 2} \ldots \ldots\left(\therefore \sec ^2 A=1+\tan ^2 A\right)$
$=\frac{\frac{1}{\cos ^2 A} \cdot \frac{\cos A}{\sin A}}{\frac{1}{\sin ^2 A}}=\frac{1}{\frac{\cos A \sin A}{\frac{1}{\sin ^2 A}}}$
$=\frac{\sin A}{\cos A}=\tan A$
View full question & answer→Question 42 Marks
Prove that : $\left(1+\frac{\cot ^2 \theta}{1+\operatorname{cosec} \theta}\right)=\operatorname{cosec} \theta$.
Answer$\text { LHS }=\left(1+\frac{\cot ^2 \theta}{1+\cos e c \theta}\right)$
$=\frac{1+\operatorname{cosec} \theta+\operatorname{cosec}{ }^2 \theta-1}{1+\operatorname{cosec} \theta}$
$=\frac{\operatorname{cosec} \theta(1+\operatorname{cosec} \theta)}{1+\operatorname{cosec} \theta}$
$=\operatorname{cosec} \theta $
$= RHS$
Hence proved.
View full question & answer→Question 52 Marks
Prove that identity:
$\frac{\sec A-1}{\sec A+1}=\frac{1-\cos A}{1+\cos A}$
Answer$\text { LHS }=\frac{\sec A-1}{\sec A+1}$
$=\frac{\frac{1}{\cos A}-1}{\frac{1}{\cos A}+1}$
$=\frac{\frac{1-\cos A}{\cos A}}{\frac{1+\cos A}{\cos A}}$
$=\frac{1-\cos A}{1+\cos A}$
$= RHS$
Hence proved.
View full question & answer→Question 62 Marks
Prove that $\frac{\tan \theta+\sin \theta}{\tan \theta-\sin \theta}=\frac{\sec \theta+1}{\sec \theta-1}$
Answer$\text { LHS }=\frac{\frac{\sin \theta}{\cos \theta}+\sin \theta}{\frac{\sin \theta}{\cos \theta}-\sin \theta}$
$=\frac{\sin \theta\left(\frac{1}{\cos \theta}+1\right)}{\sin \theta\left(\frac{1}{\cos \theta}-1\right)}$
$=\frac{\sec \theta+1}{\sec \theta-1}$
$=\frac{\sec \theta+1}{\sec \theta-1}$
$= RHS$
Hence proved.
View full question & answer→Question 72 Marks
Prove that identity:
$\frac{\sec A-1}{\sec A+1}=\frac{1-\cos A}{1+\cos A}$
Answer$\text { LHS }=\frac{\sec A-1}{\sec A+1}$
$=\frac{\frac{1}{\cos A}-1}{\frac{1}{\cos A}+1}$
$=\frac{\frac{1-\cos A}{\cos A}}{\frac{1+\cos A}{\cos A}}$
$=\frac{1-\cos A}{1+\cos A}$
$= RHS$
Hence proved.
View full question & answer→Question 82 Marks
Prove that: $\sin^4\theta + \cos^4\theta = 1 - 2sin^2\theta \cos^2\theta .$
Answer$LHS = (\sin^2\theta )^2 + (\cos^2\theta )^2 + 2 \sin^2\theta \cos^2\theta - 2 \sin^2\theta \cos^2\theta$
$= ( \sin^2\theta + \cos^2\theta )^2- 2 \sin^2\theta \cos^2\theta$
$= 1 - 2 \sin^2\theta \cos^2\theta$
$= RHS$
Hence proved.
View full question & answer→Question 92 Marks
Prove that the following identities:
Sec A( 1 + sin A)( sec A - tan A) = 1.
AnswerLHS = sec A(1 + sin A )( sec A - tan A)
$=\frac{1}{\cos A}(1+\sin A)\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}\right)$
$=\frac{1}{\cos A}(1+\sin A)\left(\frac{1-\sin A}{\cos A}\right)$
$=\frac{1-\sin ^2 A}{\cos ^2 A}=\frac{\cos ^2 A}{\cos ^2 A}$
= 1
= RHS
Hence proved.
View full question & answer→Question 102 Marks
Prove the following identities:
$\frac{1-\tan ^2 \theta}{\cot ^2 \theta-1}=\tan ^2 \theta$
Answer$\text { LHS }=\frac{1-\tan ^2 \theta}{\cot ^2 \theta-1}$
$=\frac{1-\tan ^2 \theta}{\frac{1}{\tan ^2 \theta}-1}$
$=\left(\frac{1-\tan ^2 \theta}{1-\tan ^2 \theta} / \tan ^2 \theta\right)$
$= \tan^2\theta $
$= RHS$
Hence proved.
View full question & answer→Question 112 Marks
Prove the following trigonometric identities.
$\frac{1-\cos A}{1+\cos A}=(\cot A-\operatorname{cosec} A)^2$
AnswerWe need to prove $\frac{1-\cos A}{1+\cos A}=(\cot A-\operatorname{cosec} A)^2$
Now, rationalising the L.H.S, we get
$\frac{1-\cos A}{1+\cos A}=\left(\frac{1-\cos A}{1+\cos A}\right)\left(\frac{1-\cos A}{1-\cos A}\right)$
$\left.=\frac{(1-\cos A)^2}{1-\cos ^2 A} \quad \text { (using } a^2-b^2=(a+b)(a-b)\right)$
$\left.=\frac{1+\cos ^2 A-2 \cos A}{\sin ^2 A} \quad \text { (Using } \sin ^2 \theta=1-\cos ^2 \theta\right)$
$=\frac{1}{\sin ^2 A}+\frac{\cos ^2 A}{\sin ^2 A}-\frac{2 \cos A}{\sin ^2 A}$
Using $\operatorname{cosec} \theta=\frac{1}{\sin \theta}$ and $\cot \theta=\frac{\cos \theta}{\sin \theta}$ we get
$\frac{1}{\sin ^2 A}+\frac{\cos ^2 A}{\sin ^2 A}-\frac{2 \cos A}{\sin ^2 A}=\operatorname{cosec}{ }^2 A+\cot ^2 A-2 \cot A \operatorname{cosec} A$
$(\cot A-\operatorname{cosec} A)^2 \quad\left(\text { Using }(a+b)^2=a^2+b^2+2 a b\right)$
Hence proved.
View full question & answer→Question 122 Marks
Prove the following identities:
$\frac{1}{\sin \theta+\cos \theta}+\frac{1}{\sin \theta-\cos \theta}=\frac{2 \sin \theta}{1-2 \cos ^2 \theta}$
Answer$\text { LHS }=\frac{1}{\sin \theta+\cos \theta}+\frac{1}{\sin \theta-\cos \theta}$
$=\frac{(\sin \theta-\cos \theta)+(\sin \theta+\cos \theta)}{\sin ^2 \theta-\cos ^2 \theta}$
$=\frac{2 \sin \theta}{\left(1-\cos ^2 \theta\right)-\cos ^2 \theta}$
$=\frac{2 \sin \theta}{1-2 \cos ^2 \theta}$
$= RHS$
Hence proved.
View full question & answer→Question 132 Marks
Prove the following identities:$\cot \theta - \tan \theta = \frac{2 \cos ^2 \theta-1}{\sin \theta \cos \theta}$
Answer$LHS = \cot \theta- \tan \theta$
$=\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}$
$=\frac{\cos ^2 \theta-\sin ^2 \theta}{\sin \theta \cdot \cos \theta}$
$=\frac{\cos ^2 \theta-\left(1-\cos ^2 \theta\right)}{\sin \theta \cdot \cos \theta}$
$=\frac{2 \cos ^2 \theta-1}{\sin \theta \cdot \cos \theta}$
$= RHS$
Hence proved.
View full question & answer→Question 142 Marks
Prove the following identities: $\sec^2\theta + cosec^2\theta = \sec^2\theta\ cosec^2\theta $.
AnswerLHS = $\sec^2\theta + cosec^2\theta$
$=\frac{1}{\cos ^2 \theta}+\frac{1}{\sin ^2 \theta}$
$=\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin ^2 \theta \cdot \cos ^2 \theta}$
$=\frac{1}{\sin ^2 \theta \cdot \cos ^2 \theta}$
$=\frac{1}{\sin ^2 \theta} \times \frac{1}{\cos ^2 \theta}$
$= \sec^2\theta\ cosec^2\theta$
= RHS
Hence proved.
View full question & answer→Question 152 Marks
Prove the following trigonometric identities cos theta $\frac{\cos \theta}{1-\sin \theta}=\frac{1+\sin \theta}{\cos \theta}$
AnswerWe know that, $\sin ^2 \theta+\cos ^2 \theta=1$
Multiplying both numerator and the denominator by $(1+\sin \theta)$ we have
$\frac{\cos \theta}{1-\sin \theta}=\frac{\cos \theta(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{\cos \theta(1+\sin \theta)}{1-\sin ^2 \theta}$
$=\frac{\cos \theta(1+\sin \theta)}{\cos ^2 \theta}$
$=\frac{1+\sin \theta}{\cos \theta}$
View full question & answer→Question 162 Marks
If$ A + B = 906\circ,$ show that $\frac{\sin B+\cos A}{\sin A}=2 \tan B+\tan A$.
Answer$\text { LHS }=\frac{\sin B+\sec A}{\sin A}$
$=\frac{\sin (90-A)+\sec A}{\sin A}$
$=\frac{\cos A+\sec A}{\sin A}$
$=\frac{\cos ^2 A+1}{\sin A \cdot \cos A}$
$=\frac{2 \cos ^2 A+\sin ^2 A}{\sin ^2 \cdot \cos A}$
$= 2\cot A + \tan A$
$= 2 \tan B + \tan A = RHS$
hence proved.
View full question & answer→Question 172 Marks
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
AnswerLHS = cos 1°cos 2°cos 3° ....cos 180°
= cos 1°cos 2°cos 3° ....cos 89° cos 90° .... cos 180°
= cos 1°cos 2°cos 3° ....cos 89° x 0 x cos 91° .... cos 180°
= 0
= RHS
Hence proved.
View full question & answer→Question 182 Marks
Without using the trigonometric table, prove that
tan 10° tan 15° tan 75° tan 80° = 1
AnswerL.H.S. = tan 10° tan 15° tan 75° tan 80°
= tan 10° tan 15° tan (90° - 15°) tan (90° - 10°)
= = tan 10° tan 15° cot 15° cot 10°
$=\frac{1}{\cot 10}^{\circ} \times \frac{1}{\cot 15}^{\circ} \times \cot 15^{\circ} \times \cot 10^{\circ}$
= 1
= RHS
Hence proved.
View full question & answer→Question 192 Marks
Proved that $cosec^2(90^\circ - \theta ) - \tan^2 \theta = \cos^2(90^\circ - \theta ) + \cos^2 \theta .$
Answer$LHS = cosec^2(90^\circ - \theta ) - \tan^2 \theta$
$LHS = sec^2\theta - \tan^2 \theta = 1$
$RHS = \cos^2(90^\circ - \theta ) + \cos^2 \theta$
$RHS = \sin^2\theta + \cos^2 \theta = 1$
Hence, $LHS = RHS$
Hence proved.
View full question & answer→Question 202 Marks
Prove that $\frac{\sin \theta \cdot \cos \left(90^{\circ}-\theta\right) \cos \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\cos \theta \sin \left(90^{\circ}-\theta\right) \sin \theta}{\cos \left(90^{\circ}-\theta\right)}=1$
Answer$\text { LHS }=\frac{\sin \theta \cdot \cos \left(90^{\circ}-\theta\right) \cos \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\cos \theta \sin \left(90^{\circ}-\theta\right) \sin \theta}{\cos \left(90^{\circ}-\theta\right)}$
$=\frac{\sin \theta \cdot \sin \theta \cos \theta}{\cos \theta}+\frac{\cos \theta \cdot \cos \theta \sin \theta}{\sin \theta}$
$=\sin^2 \theta +\cos^2\theta $
$= 1$
= RHS
Hence proved.
View full question & answer→Question 212 Marks
Prove that $\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\sin \theta}{\cos \left(90^{\circ}-\theta\right)}=2$
Answer$\text { LHS }=\frac{\cos \theta}{\sin \left(90^{\circ}-\theta\right)}+\frac{\sin \theta}{\cos \left(90^{\circ}-\theta\right)}$
$=\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\sin \theta}$
$= 1 + 1 = 2$
$= RHS$
Hence proved.
View full question & answer→Question 222 Marks
Prove that $\cos \theta \sin (90^\circ - \theta ) + \sin \theta \cos (90^\circ - \theta ) = 1.$
Answer$LHS = \cos \theta \sin (90^\circ - \theta ) + \sin \theta \cos (90^\circ - \theta )$
$= \cos \theta . \cos \theta + \sin \theta . \sin \theta$
$= \cos^2\theta + \sin ^2\theta$
$= 1$
$= RHS$
Hence proved.
View full question & answer→Question 232 Marks
Prove that $\sin^2\theta + \cos^4 \theta = \cos^2\theta + \sin^4 \theta .$
Answer$L.H.S. = \sin^2\theta + \cos^4 \theta$
$= 1 - \cos^2 \theta + \cos^4 \theta$
$= 1 - \cos^2 \theta (1 - \cos^2 \theta )$
$= 1 - (1 - \sin^2 \theta ) \sin^2 \theta$
$= 1 - \sin^2 \theta + \sin^4 \theta$
$= \cos^2 \theta + \sin^4 \theta$
$= R.H.S.$
Hence proved.
View full question & answer→Question 242 Marks
If $A, B, C$ are the interior angles of a triangle $ABC,$ prove that $\tan \frac{B+C}{2}=\cot \frac{A}{2}$
AnswerIn $\triangle ABC,$ we have
$A + B + C = 180^\circ$
$\Rightarrow B + C = 180^\circ – A$
$\Rightarrow \frac{B+C}{2}=90^{\circ}-\frac{A}{2}$
Taking tan on both sides, we get
$ \Rightarrow \tan \left(\frac{B+C}{2}\right)=\tan \left(90^{\circ}-\frac{A}{2}\right)$
$\Rightarrow \tan \left(\frac{B+C}{2}\right)=\cot \frac{A}{2} $
View full question & answer→Question 252 Marks
Prove that $\frac{\sin \left(90^{\circ}-\theta\right)}{\cos \theta}+\frac{\tan \left(90^{\circ}-\theta\right)}{\cot \theta}+\frac{\operatorname{cosec}\left(90^{\circ}-\theta\right)}{\sec \theta}=3$.
AnswerLHS $=\frac{\sin \left(90^{\circ}-\theta\right)}{\cos \theta}+\frac{\tan \left(90^{\circ}-\theta\right)}{\cot \theta}+\frac{\operatorname{cosec}\left(90^{\circ}-\theta\right)}{\sec \theta}$
$=\frac{\cos \theta}{\cos \theta}+\frac{\cot \theta}{\cot \theta}+\frac{\sec \theta}{\sec \theta}$
= 1 + 1 + 1
= 3
= RHS
Hence proved.
View full question & answer→Question 262 Marks
Prove that sin $(90^\circ - \theta ) \cos (90^\circ - \theta ) = \tan \theta . \cos^2\theta .$
Answer$LHS = \sin (90^\circ - \theta ) \cos (90^\circ - \theta )$
$LHS = \cos \theta . \sin \theta$
$RHS = \tan \theta . \cos^2\theta$
RHS = $\frac{\sin \theta}{\cos \theta} \times \cos ^2 \theta$
$RHS = \cos \theta . \sin \theta$
$\therefore LHS = RHS$
Hence proved.
View full question & answer→Question 272 Marks
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
AnswerLHS = sin θ sin( 90° - θ) - cos θ cos( 90° - θ)
= sin θ . cos θ - cos θ . sin θ
= 0
= RHS
Hence proved.
View full question & answer→Question 282 Marks
Prove that $\sin( 90^\circ - \theta ) \sin \theta \cot \theta = \cos^2\theta .$
Answer$LHS = \sin( 90^\circ - \theta ) \sin \theta \cot \theta$
$= \cos \theta . \sin \theta $.$\frac{\cos \theta}{\sin \theta}$
$= \cos^2\theta$
$= RHS$
Hence proved.
View full question & answer→Question 292 Marks
Prove that $\frac{\tan \theta}{\cot \left(90^{\circ}-\theta\right)}+\frac{\sec \left(90^{\circ}-\theta\right) \sin \left(90^{\circ}-\theta\right)}{\cos \theta \cdot \cos e c} \theta=2$.
AnswerLHS $=\frac{\tan \theta}{\cot \left(90^{\circ}-\theta\right)}+\frac{\sec \left(90^{\circ}-\theta\right) \sin \left(90^{\circ}-\theta\right)}{\cos \theta \cdot \operatorname{cosec} \theta}$
$=\frac{\tan \theta}{\tan \theta}+\frac{\operatorname{cosec} \theta \cdot \cos \theta}{\cos \theta \cdot \operatorname{cosec} \theta}$
= 1 + 1 = 2
= RHS
Hence proved.
View full question & answer→Question 302 Marks
Prove that $cosec^2 (90^\circ - \theta ) + \cot^2 (90^\circ - \theta ) = 1 + 2 \tan^2 \theta .$
Answer$LHS = cosec^2 (90^\circ - \theta ) + \cot^2 (90^\circ - \theta )$
$= sec^2 \theta + \tan^2\theta$
$= 1 + \tan^2\theta + \tan^2\theta$
$= 1 + 2 \tan^2\theta$
$= RHS$
Hence proved.
View full question & answer→Question 312 Marks
Prove that $\sec^2 (90^\circ - \theta ) + \tan^2 (90^\circ - \theta ) = 1 + 2 \cot^2 \theta .$
Answer$LHS = \sec^2 (90^\circ - \theta ) + \tan^2 (90^\circ - \theta )$
$= cosec^2\theta + \cot^2\theta $
$= 1 + \cot^2\theta + \cot^2\theta $
$= 1 + 2cot^2\theta $
$= RHS$
Hence proved.
View full question & answer→Question 322 Marks
Prove that $sec \theta . cosec (90^\circ - \theta ) - \tan \theta . \cot( 90^\circ - \theta ) = 1.$
Answer$LHS = sec \theta . cosec (90^\circ - \theta ) - \tan \theta . \cot( 90^\circ - \theta )= sec \theta . sec \theta - \tan \theta . \tan \theta$
$= sec^2\theta - \tan^2\theta$
$= 1$
$= RHS$
Hence proved.
View full question & answer→Question 332 Marks
Prove that $\cot \theta . \tan (90^\circ - \theta ) - \sec (90^\circ - \theta ). cosec \theta + 1 = 0$.
Answer$LHS = \cot \theta . \tan (90^\circ - \theta ) - \sec (90^\circ - \theta ). cosec \theta + 1= \cot \theta . \cot \theta - cosec \theta . cosec \theta + 1$
$= (\cot^2\theta - cosec^2\theta ) + 1$
$= - 1 + 1 = 0$
$= RHS$
Hence proved.
View full question & answer→Question 342 Marks
Prove that $\frac{\cos \theta}{1-\sin \theta}=\frac{1+\sin \theta}{\cos \theta}$.
Answer$\text { L.H.S. }=\frac{\cos \theta}{1-\sin \theta}$
$=\frac{\cos \theta(1+\sin \theta)}{(1-\sin \theta)(1+\sin \theta)}$
$=\frac{\cos \theta(1+\sin \theta)}{1-\sin ^2 \theta}$
$=\frac{\cos \theta(1+\sin \theta)}{\cos ^2 \theta}$
$=\frac{1+\sin \theta}{\cos \theta}$
Hence proved.
View full question & answer→Question 352 Marks
Prove that : $\frac{\sin \left(90^{\circ}-\theta\right) \tan \left(90^{\circ}-\theta\right) \sec \left(90^{\circ}-\theta\right)}{\cos e c \theta \cdot \cos \theta \cdot \cot \theta}=1$
AnswerLHS = $\frac{\sin \left(90^{\circ}-\theta\right) \tan \left(90^{\circ}-\theta\right) \sec \left(90^{\circ}-\theta\right)}{\operatorname{cosec} \theta \cdot \cos \theta \cdot \cot \theta}=1$
$ =\frac{\operatorname{cosec} \theta \cdot \cos \theta \cdot \cot \theta}{\operatorname{cosec} \theta \cdot \cos \theta \cdot \cot \theta}$
$=1$
$=\text { RHS } $
Hence proved.
View full question & answer→Question 362 Marks
If $A = 30^\circ,$ verify that $\sin 2 A=\frac{2 \tan A}{1+\tan ^2 A}$
Answer$LHS =\sin 2A$
Putting $A = 30^\circ$ in $LHS$ and $RHS.,$ we get
LHS = sin 2 x 30° = sin 60° $=\frac{\sqrt{3}}{2}$
$\text { RHS }=\frac{2 \times \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}=\frac{2 \times \frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^2}$
$=\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}} \cdot \frac{2}{\frac{4}{3}}$
$=\frac{2 \times 3}{\sqrt{3} \times 4}=\frac{\sqrt{3}}{4}$
Hence,
$LHS = RHS$
Hence proved.
View full question & answer→Question 372 Marks
Prove that : $1-\frac{\cos ^2 \theta}{1+\sin \theta}=\sin \theta$
Answer$\text { LHS }=1-\frac{\cos ^2 \theta}{1+\sin \theta} $
$ =1-\frac{1-\sin ^2 \theta}{1+\sin \theta} $
$ =1-\frac{(1-\sin \theta)(1+\sin \theta)}{1+\sin \theta}$
$= 1 - ( 1 - \sin \theta )$
$= 1 - 1 + \sin \theta$
$= \sin \theta$
$= RHS$
Hence proved.
View full question & answer→Question 382 Marks
If $x = a \sec \theta+ b \tan \theta$ and $y = a \tan \theta+ b \sec \theta$ prove that $x^2-y^2=a^2-b^2$.
AnswerHere,
$x^2= a^2 sec^2\theta + 2ab sec \theta .\tan \theta + b^2tan^2\theta$
$y^2= a^2tan^2\theta + 2ab sec \theta .\tan \theta + b^2sec^2\theta$
$\Rightarrow x^2- y^2 = a^2 ( sec^2\theta - ^\tan^2\theta ) - b^2 ( sec^2\theta - ^\tan^2\theta )$
$\Rightarrow x^2 - y^2 = a^2- b^2.....( \because sec^2\theta - ^\tan^2\theta = 1)$
Hence proved.
View full question & answer→Question 392 Marks
Prove that $( 1 + \tan A)^2 + (1 - \tan A)^2 = 2 \sec^2A$
Answer$LHS = ( 1 + \tan A)^2 + (1 - \tan A)^2$
$= 1 + 2 \tan A + \tan^2A + 1 - 2 \tan A + \tan^2A$
$= 2( 1 + \tan^2A)$
$= 2 \sec^2A$
$= RHS$
Hence proved.
View full question & answer→Question 402 Marks
Prove that: $\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}=1-2 \sec \theta \cdot \tan \theta+2 \tan ^2 \theta$
Answer$\text { LHS }=\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}$
$=\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta} \times \frac{\sec \theta-\tan \theta}{\sec \theta-\tan \theta}$
$=\frac{(\sec \theta-\tan \theta)^2}{\sec ^2 \theta-\tan ^2 \theta}$
$=\frac{\sec ^2 \theta+\tan ^2 \theta-2 \sec \theta \cdot \tan \theta}{1}$
$= 1 + 2 \tan^2\theta - 2 \sec \theta. \tan \theta$
$= R.H.S.$
Hence proved.
View full question & answer→Question 412 Marks
Without using trigonometric tables, find the value of (sin 72° + cos 18°)(sin 72° - cos 18°).
Answer(sin 72° + cos 18°)(sin 72° - cos 18°)
= ( sin (90° - 18°) + cos 18° )( sin(90° - 18°) - cos 18° )
= ( cos 18° + cos 18°)(cos 18° - cos 18°)
= 0
View full question & answer→Question 422 Marks
Without using tables evaluate: 3cos 80°. cosec 10° + 2sin 59° sec 31°
Answer3 cos 80°. cosec 10° + 2 sin 59° sec 31°
= 3cos (90° – 10°) cosec 10° + 2sin (90° – 31°) sec 31°
= 3 sin 10° cosec 10° + 2 cos 31°, sec 31°....(sin (90° – θ) = cos θ, cos (90° – θ) = sin θ)
= 3 × 1 + 2 × 1 .....(∵ sin θ. Cosec θ = 1, cos θ. sec θ = 1)
= 5
View full question & answer→Question 432 Marks
$\frac{\sin 20^{\circ} 50 \prime+\tan 67^{\circ} 40 \prime}{\cos 32^{\circ} 20 \prime-\sin 15^{\circ} 10 \prime}$
Answer$\frac{\sin 20^{\circ} 50 \prime+\tan 67^{\circ} 40 \prime}{\cos 32^{\circ} 20 \prime-\sin 15^{\circ} 10 \prime}$
$=\frac{0.3557+2.4340}{0.8450-2.616} $
$ =\frac{2.7897}{0.5834} $
$=\frac{27897}{5834}$
$ =4.7818 .$
View full question & answer→Question 442 Marks
$\frac{\sin 40^{\circ}+\cos 50^{\circ}}{\tan 38^{\circ} 20 \prime}$
Answer$\frac{\sin 40^{\circ}+\cos 50^{\circ}}{\tan 38^{\circ} 20 \prime} $
$=\frac{0.6428+0.6428}{0.7907}$
$ =\frac{1.2856}{0.7907}$
$ =1.6259 .$
View full question & answer→Question 452 Marks
The string of a kite is 150 m long and it makes an angle of 60° with the horizontal. Find the height of the kite from the ground.
AnswerLet h be the height of the kite.
$PB$ be the length of string such that $PB = 150\ m.$
In right-angled $\triangle BPA,$
$\sin 60^\circ = \frac{h}{150}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{h}{150}$
$\Rightarrow h =\frac{150 \sqrt{3}}{2}$
$\Rightarrow h =75 \sqrt{3}$
$h = 1.732 \times 75$
$h = 129.9 m$
Hence, the height of kite above the ground $= 129.9\ m.$
View full question & answer→Question 462 Marks
Solve: $2 \cos ^2 \theta+\sin \theta-2=0$.
Answer$2 \cos 2 \theta+\sin \theta-2=0$
$\Rightarrow 2\left(1-\sin ^2 \theta\right)+\sin \theta-2=0$
$\Rightarrow 2-2 \sin ^2 \theta+\sin \theta-2=0$
$\Rightarrow-\sin \theta(2 \sin \theta-1)=0$
$\Rightarrow \sin \theta(2 \sin \theta-1)=0$
$\Rightarrow \sin \theta=0 \text { or } 2 \sin \theta-1=0$
$\Rightarrow \sin \theta=0 \text { or } \sin \theta=\frac{1}{2}$
$\Rightarrow \theta=30^{\circ}$
View full question & answer→