# Prove the following identities,

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) $\frac{\tan \theta}{\mathbf{1}-\cot \theta}+\frac{\cot \theta}{\mathbf{1}-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$

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

(v) $\frac{\cos \mathbf{A}-\sin \mathbf{A}+\mathbf{1}}{\cos \mathbf{A}+\sin \mathbf{A}-\mathbf{1}}=\operatorname{cosec} \mathrm{A}+\cot \mathrm{A}$, using the identity $\operatorname{cosec}^{2} \mathrm{~A}=1+\cot ^{2} \mathrm{~A}$.

(vi) $\sqrt{\frac{\mathbf{1}+\sin \mathbf{A}}{\mathbf{1}-\sin \mathbf{A}}}=\sec \mathbf{A}+\tan \mathbf{A} .$

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

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

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

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

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

(x) $\left(\frac{1+\tan ^{2} \mathbf{A}}{\mathbf{1}+\cot ^{2} \mathbf{A}}\right)=\left(\frac{\mathbf{1}-\tan \mathbf{A}}{\mathbf{1}-\cot \mathbf{A}}\right)^{2}=\tan ^{2} \mathrm{~A} .$

Solution:

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

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

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

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

$\therefore \quad \mathrm{LHS}=\mathrm{RHS} .$

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

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

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

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

$=2 \sec A=R . H . S$

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

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

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

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

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

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

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

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

$\left\{\because a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right.$

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

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

$=1+\left(\frac{1}{\cos \theta}\right)\left(\frac{1}{\sin \theta}\right)$

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

$\therefore \quad$ LHS $=$ RHS

(iv) L.H.S. $=\frac{1+\sec \mathbf{A}}{\sec \mathbf{A}}=\frac{1+\frac{1}{\cos \mathbf{A}}}{\frac{1}{\cos \mathbf{A}}}=\frac{\cos \mathbf{A}+1}{1}$

R.H.S. $=\frac{\sin ^{2} \mathbf{A}}{\mathbf{1}-\cos \mathbf{A}}=\frac{1-\cos ^{2} \mathbf{A}}{1-\cos \mathbf{A}}=1+\cos \mathbf{A}$

$\therefore$ L.H.S. $=$ R.H.S.

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

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

(Dividing the numerator and denominator by $\sin \mathrm{A})$

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

$=\frac{(\operatorname{cosec} A+\cot A)-\left(\operatorname{cosec}^{2} A-\cot ^{2} A\right)}{\{1+\cot A-\operatorname{cosec} A\}}$

$\left(\because \operatorname{cosec}^{2} \mathrm{~A}=1+\cot ^{2} \mathrm{~A}\right.$, i.e., $\left.\operatorname{cosec}^{2} \mathrm{~A}-\cot ^{2} \mathrm{~A}=1\right)$

$=\frac{(\operatorname{cosec} A+\cot A)-(\operatorname{cosec} A+\cot A) \times(\operatorname{cosec} A-\cot A)}{\{1+\cot A-\operatorname{cosec} A\}}$

$\left\{\because(a+b)(a-b)=a^{2}-b^{2}\right\}$

$=\frac{(\operatorname{cosec} A+\cot A) \times\{1-(\operatorname{cosec} A-\cot A)\}}{\{1+\cot A-\operatorname{cosec} A\}}$

$=\frac{(\operatorname{cosec} A+\cot A) \times\{1+\cot A-\operatorname{cosec} A\}}{\{1+\cot A-\operatorname{cosec} A\}}$

= cosec A + cot A

= RHS

(vi) $L H S=\sqrt{\frac{1+\sin \mathbf{A}}{1-\sin \mathbf{A}}}=\sqrt{\frac{(1+\sin \mathbf{A})(1+\sin \mathbf{A})}{(1-\sin \mathbf{A})(1+\sin \mathbf{A})}}$

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

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

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

$\therefore \quad$ LHS $=$ RHS .

(vii) L.H.S. $=\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}$

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

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

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

$=\tan \theta=\mathrm{R} . \mathrm{H} . \mathrm{S}$

(viii) L.H.S. $=(\sin \mathrm{A}+\operatorname{cosec} \mathrm{A})^{2}+(\cos \mathrm{A}+\sec \mathrm{A})^{2}$

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

$=4+1+1+\cot ^{2} \mathrm{~A}+1+\tan ^{2} \mathrm{~A}$

$=7+\tan ^{2} A+\cot ^{2} A=$ R.H.S.

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

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

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

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

Now, RHS $=\frac{1}{\tan \boldsymbol{A}+\cot \boldsymbol{A}}=\frac{1}{\frac{\sin \boldsymbol{A}}{\cos \boldsymbol{A}}+\frac{\cos \boldsymbol{A}}{\sin \boldsymbol{A}}}$

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

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

$\therefore \quad$ LHS $=$ RHS .

(x) L.H.S. $=\left(\frac{1+\operatorname{tn}^{2} \mathbf{A}}{\mathbf{1}+\cot ^{2} \mathbf{A}}\right)=\frac{\sec ^{2} \mathbf{A}}{\operatorname{cosec}^{2} \mathbf{A}}$

$=\tan ^{2} \mathrm{~A}=\mathrm{R} . \mathrm{H} . \mathrm{S}$

$\&\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\left(\frac{\frac{\cos A-\sin A}{\cos A}}{\frac{\sin A-\cos A}{\sin A}}\right)^{2}$

$=\frac{\sin ^{2} A}{\cos ^{2} A}=\tan ^{2} A=$ R.H.S.