Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) $ (\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta} $
(ii) $ \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A $
(iii) $ \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta $
[Hint: Write the expression in terms of sin θ and cos θ]
(iv) $ \frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A} $
[Hint: Simplify LHS and RHS separately]
(v) $ \frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A, u s i n g $ the identity $ \operatorname{cosec}^{2} A=1+\cot ^{2} A $.
(vi) $ \sqrt{\frac{1+\sin \mathrm{A}}{1-\sin \mathrm{A}}}=\sec \mathrm{A}+\tan \mathrm{A} $
(vii) $ \frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta $
(viii) $ (\sin \mathrm{A}+\operatorname{cosec} \mathrm{A})^{2}+(\cos \mathrm{A}+\sec \mathrm{A})^{2}=7+\tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A} $
(ix) $ (\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A} $
[Hint : Simplify LHS and RHS separately]
(x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A.

AcademicMathematicsNCERTClass 10

To do:

We have to prove the given identities.

Solution:  

(i) LHS $=(\operatorname{cosec} \theta-\cot \theta)^{2}$

$=(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta})^{2}$

$=(\frac{1-\cos \theta}{\sin \theta})^{2}$

$=\frac{(1-\cos \theta)(1-\cos \theta)}{\sin ^{2} \theta}$

$=\frac{(1-\cos \theta)(1-\cos \theta)}{(1-\cos ^{2} \theta)}$

$=\frac{(1-\cos \theta)(1-\cos \theta)}{(1+\cos \theta)(1-\cos \theta)}$

$=\frac{1-\cos \theta}{1+\cos \theta}$

$=$ RHS 

Hence proved.

(ii) LHS $=\frac{\cos \mathrm{A}}{1+\sin \mathrm{A}}+\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}}$

$=\frac{(\cos \mathrm{A})^{2}+(1+\sin \mathrm{A})^{2}}{\cos \mathrm{A}(1+\sin \mathrm{A})}$

$=\frac{\cos ^{2} \mathrm{~A}+1+\sin ^{2} \mathrm{~A}+2 \sin \mathrm{A}}{\cos \mathrm{A}(1+\sin \mathrm{A})}$

$=\frac{1+1+2 \sin \mathrm{A}}{\cos \mathrm{A}(1+\sin \mathrm{A})}$

$=\frac{2+2 \sin \mathrm{A}}{\cos \mathrm{A}(1+\sin \mathrm{A})}$

$=\frac{2(1+\sin \mathrm{A})}{\cos \mathrm{A}(1+\sin \mathrm{A})}$

$=\frac{2}{\cos \mathrm{A}}$

$=2 \sec \mathrm{A}$

$=$ RHS

Hence proved.

(iii) LHS $=\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$

$=\frac{\frac{\sin \theta}{\cos \theta}}{\frac{1}{1}-\frac{\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{\frac{1}{1}-\frac{\sin \theta}{\cos \theta}}$

$=\frac{\sin \theta}{\cos \theta} \times(\frac{\sin \theta}{\sin \theta-\cos \theta})+\frac{\cos \theta}{\sin \theta} \times(\frac{\cos \theta}{\cos \theta-\sin \theta})$

$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}-\frac{\cos ^{2} \theta}{\sin \theta(\sin \theta-\cos \theta)}$

$=\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta \cos \theta(\sin \theta-\cos \theta)}$

$=\frac{(\sin \theta-\cos \theta)(\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \cos \theta)}{\sin \theta \cos \theta(\sin \theta-\cos \theta)}$

$=\frac{\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \cos \theta}{\sin \theta \cos \theta}$

$=\frac{1+\sin \theta \cos \theta}{\sin \theta \cos \theta}$

$=\frac{1}{\sin \theta \cos \theta}+\frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$

$=\sec \theta \operatorname{cosec} \theta+1$

$=$ RHS

Hence proved. 

(iv) LHS $=\frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}$

$=\frac{\frac{1}{1}+\frac{1}{\cos \mathrm{A}}}{\frac{1}{\cos \mathrm{A}}}$

$=(\frac{\cos \mathrm{A}+1}{\cos \mathrm{A}}) \times \frac{\cos \mathrm{A}}{1}$

$=\cos \mathrm{A}+1$

RHS $=\frac{\sin ^{2} \mathrm{~A}}{1-\cos \mathrm{A}}$

$=\frac{1-\cos ^{2} \mathrm{~A}}{1-\cos \mathrm{A}}$

$=\frac{(1+\cos \mathrm{A})(1-\cos \mathrm{A})}{1-\cos \mathrm{A}}$

$=1+\cos \mathrm{A}$

LHS $=$ RHS

Hence proved.  

(v) LHS $=\frac{\cos A-\sin A+1}{\cos A+\sin A-1}$

Dividing numerator and denominator by $\sin \mathrm{A}$, we get,

$=\frac{\frac{\cos A}{\sin A}-\frac{\sin A}{\sin A}+\frac{1}{\sin A}}{\frac{\cos A}{\sin A}+\frac{\sin A}{\sin A}-\frac{1}{\sin A}}$

$=\frac{\cot \mathrm{A}-1+\operatorname{cosec} \mathrm{A}}{\cot \mathrm{A}+1-\operatorname{cosec} \mathrm{A}}$

$=\frac{\operatorname{cosec} \mathrm{A}+\cot \mathrm{A}-1}{1+\cot \mathrm{A}-\operatorname{cosec} \mathrm{A}}$

$=\frac{\operatorname{cosec} \mathrm{A}+\cot \mathrm{A}-(\operatorname{cosec}^{2} \mathrm{~A}-\cot ^{2} \mathrm{~A})}{1+\cot \mathrm{A}-\operatorname{cosec} \mathrm{A}}$                 (Since $\operatorname{cosec}^{2} \mathrm{~A}-\cot ^{2} \mathrm{~A}=1$)

$=\frac{\operatorname{cosec} \mathrm{A}+\cot \mathrm{A}-[(\operatorname{cosec} \mathrm{A}+\cot \mathrm{A})(\operatorname{cosec} \mathrm{A}-\cot \mathrm{A})]}{1+\cot \mathrm{A}-\operatorname{cosec} \mathrm{A}}$

$=\frac{(\operatorname{cosec} \mathrm{A}+\cot \mathrm{A})[1-\operatorname{cosec} \mathrm{A}+\cot \mathrm{A}]}{1+\cot \mathrm{A}-\operatorname{cosec} \mathrm{A}}$

$=\operatorname{cosec} \mathrm{A}+\cot \mathrm{A}$

$=$ RHS

LHS $=$ RHS

Hence proved.

(vi)  LHS $=\sqrt{\frac{1+\sin \mathrm{A}}{1-\sin \mathrm{A}}}$

$=\sqrt{\frac{(1+\sin \mathrm{A})}{(1-\sin \mathrm{A})} \times \frac{(1+\sin \mathrm{A})}{(1+\sin \mathrm{A})}}$

$=\sqrt{\frac{(1+\sin \mathrm{A})^{2}}{(1)^{2}-(\sin \mathrm{A})^{2}}}$

$=\sqrt{\frac{(1+\sin \mathrm{A})^{2}}{\cos ^{2} \mathrm{~A}}}$

$=\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}}$

$=\frac{1}{\cos \mathrm{A}}+\frac{\sin \mathrm{A}}{\cos \mathrm{A}}$

$=\sec \mathrm{A}+\tan \mathrm{A}$

$=$ RHS

LHS $=$ RHS

Hence proved. 

(vii) LHS $=\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}$

$=\frac{\sin \theta(1-2 \sin ^{2} \theta)}{\cos \theta(2 \cos ^{2} \theta-1)}$

$=\frac{\sin \theta}{\cos \theta} \times \frac{(1-2(1-\cos ^{2} \theta)}{(2 \cos ^{2} \theta-1)}$

$=\frac{\sin \theta}{\cos \theta} \times \frac{(1-2+2 \cos ^{2} \theta)}{(2 \cos ^{2} \theta-1)}$

$=\frac{\sin \theta}{\cos \theta} \times \frac{(-1+2 \cos ^{2} \theta)}{(2 \cos ^{2} \theta-1)}$

$=\frac{\sin \theta}{\cos \theta} \times \frac{(2 \cos ^{2} \theta-1)}{(2 \cos ^{2} \theta-1)}$

$=\frac{\sin \theta}{\cos \theta}$

$=\tan \theta$

$=$ RHS

LHS $=$ RHS

Hence proved. 

(viii) LHS $=(\sin A+\operatorname{cosec} A)^{2}+(\cos A+\sec A)^{2}$

$=\sin ^{2} A+2 \sin A \operatorname{cosec} A+\operatorname{cosec}^{2} A+\cos ^{2} A+2\cos A \sec A+\sec ^{2} A$

$=\sin ^{2} A+2\sin A \frac{1}{\sin A}+\operatorname{cosec}^{2} A+\cos ^{2} A+2\cos A\frac{1}{\cos A}+\sec ^{2} A$

$=\sin ^{2} A+2+\operatorname{cosec}^{2} A+\cos ^{2} A+2+\sec ^{2} A$

$=\sin ^{2} A+\cos ^{2} A+2+2+\operatorname{cosec}^{2} A+\sec ^{2} A$

$=1+4+(1+\tan ^{2} A)+(1+\cot ^{2} A)$

$=7+\tan ^{2} A+\cot ^{2} A$

$=$ RHS

LHS $=$ RHS

Hence proved.  

(ix) LHS $=(\operatorname{cosec} A-\sin A)(\sec A-\cos A)$

$=(\frac{1}{\sin A}-\sin A)(\frac{1}{\cos A}-\cos A)$

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

$=\frac{\cos ^{2} A}{\sin A} \times \frac{\sin ^{2} A}{\cos A}$

$=\sin A \cos A$

RHS $=\frac{1}{\tan A+\cot A}$

$=\frac{1}{\frac{\cos A}{\sin A}+\frac{\sin A}{\cos A}}$

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

$=\frac{1}{\frac{1}{\sin A \cos A}}$

$=\frac{1}{1} \times \frac{\sin A \cos A}{1}$

$=\sin A \cos A$

LHS $=$ RHS

Hence proved.  

raja
Updated on 10-Oct-2022 13:23:27

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