Question

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i) ${(\text{cosec}\theta -\cot \theta )}^2=\frac{1-\cos \theta }{1+\cos \theta }$

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

(iii) $\frac{\tan \theta }{1-\cot \theta }+\frac{\cot \theta }{1-\tan \theta }=1+\sec \theta \text{cosec}\theta$[Hint: Write the expression in terms of $\sin \theta$ and $\cos \theta$

(iv) $\frac{1+\sec A}{\sec A}=\frac{{\sin }^{2}A}{1-\cos A}$ [Hint: Simplify LHS and RHS separately].

(v) $\frac{\cos A-\sin A-1}{\cos A+\sin A+1}=\text{cosec}A+\cot A$, Using the identity ${\text{cosec}}^{2}A=1+{\cot }^{2}A$

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

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

(viii) ${(\sin A+\text{cosec}A)}^2+{(\cos A+\sec A)}^2=7+{\tan }^{2}A+{\cot }^{2}A$

(ix) $(\text{cosec}A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$[Hint: Simplify LHS and RHS separately]

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

Answer 1

(i)

LHS $=(\text{cosec}\theta -\cot \theta {)}^2={(\frac{1}{\sin \theta }-\frac{\cos \theta }{\sin \theta })}^2=\frac{(1-\cos \theta {)}^2}{{\sin }^2\theta }$

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

$=RHS$

(ii)

LHS : $\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=\frac{{\cos }^{2}A+(1+\sin A{)}^2}{(1+\sin A)\left(\cos A\right)}$

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

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

$=RHS$

(iii)

LHS : $\frac{\tan \theta }{1-\cos \theta }+\frac{\cot \theta }{1-\tan \theta }$

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

$=\frac{1}{\sin \theta -\cos \theta }[\frac{{\sin }^2\theta }{\cos \theta }-\frac{{\cos }^2\theta }{\sin \theta }]$

$=\left[\frac{1}{\sin \theta -\cos \theta }\right]\left[\frac{{\sin }^3\theta -{\cos }^3\theta }{\sin \theta \cos \theta }\right]$

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

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

$=$RHS

(iv)

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

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

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

$=$RHS

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

$=\frac{\cot A-1+\text{cosec}A}{\cos A+1-\text{cosec}A}$

$\frac{(\cot A-1+\text{cosec}A{)}^2}{(\cot A{)}^2-(1-\text{cosec}A{)}^2}$

$=\frac{(2\text{cosec}A-2)(\text{cosec}A+\cot A)}{(-1-1+2\cos A)}$

$=\text{cosec}A+\cot A$

$=RHS$

(vi)

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

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

$=\sec A+\tan A$

$=RHS$

(vii)

$\frac{\sin \theta -2{\sin }^3\theta }{2{\cos }^3\theta -\cos \theta }=\frac{\sin \theta (1-2{\sin }^2\theta )}{\cos \theta (2{\cos }^2\theta -1)}$

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

$=\tan \theta =$ RHS

(viii)

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

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

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

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

$=RHS$

(ix)

LHS : $(\text{cosec}A-\sin A)(\sec A-\cos A)$

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

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

$=\sin A\cos A$

RHS : $\frac{1}{\tan A+\cot A}=\frac{\sin A\cos A}{{\sin }^2+{\cos }^{2}A}=\sin A\cos A$

$\therefore LHS=RHS$

(x)

LHS : $\frac{1+{\tan }^{2}A}{1+{\cot }^{2}A}=\frac{1+\frac{{\sin }^{2}A}{{\cos }^{2}A}}{1+\frac{{\cos }^{2}A}{{\sin }^{2}A}}=\frac{\frac{1}{{\cos }^{2}A}}{\frac{1}{{\sin }^{2}A}}=\frac{{\sin }^{2}A}{{\cos }^{2}A}={\tan }^{2}A$

RHS : ${\left(\frac{1-\tan A}{1-\cot A}\right)}^2=\frac{1+{\tan }^2-2\tan A}{1+{\cot }^{2}A-2\cot A}$

$=\frac{{\sec }^{2}A-2\tan A}{{\text{cosec}}^{2}A-2\cot A}$

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

$\therefore LHS=RHS$