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

(i) (cosec θ – cot θ)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 θ cosec θ

     [Hint: Write the expression in terms of sin θ and cos θ]

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

     [Hint: Simplify L.H.S. and R.H.S. separately]

(v) $\frac{\cos A-\sin A+1}{\cos A+\sin A-1}$= cosec A + cot A, using the identity cosec²A = 1+cot²A.

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

(vii) $\frac{(\sin \theta -2{\sin }^3\theta )}{(2{\cos }^3\theta -\cos \theta )}$ = tan θ
(viii) (sin A + cosec A)² + (cos A + sec A)² = 7+tan²A+cot²A
(ix) (cosec A – sin A)(sec A – cos A) = $\frac{1}{\tan A+cotA}$
[Hint: Simplify L.H.S. and R.H.S. separately] 

(x) $\frac{1+{\tan }^{2}A}{1+co{t}^{2}A}$ = ${\left(\frac{1-\tan A}{1-cotA}\right)}^2$ = tan²A

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

L.H.S. = (cosec θ – cot θ)²

= (cosec²θ + cot²θ – 2cosec θ cot θ)

Now, apply the corresponding inverse functions and equivalent ratios to simplify

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

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

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

= $\frac{{(1-\cos \theta )}^{ }}{(1+\cos \theta )}$ = R.H.S.

Hence proved.

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

L.H.S. =  $\frac{\cos A}{1+\sin A}$ + $\frac{1+\sin A}{\cos A}$

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

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

Since cos²A + sin²A = 1, we can write it as

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

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

=$\frac{2}{\cos A}$ = 2 sec A = R.H.S.

Hence proved.

(iii)$\frac{\tan \theta }{1-cot\theta }$ + $\frac{cot\theta }{1-\tan \theta }$= 1 + sec θ cosec θ

L.H.S. =$\frac{\tan \theta }{1-cot\theta }$ + $\frac{cot\theta }{1-\tan \theta }$

We know that tan θ =$\frac{\sin \theta }{\cos \theta }$

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

Now, substitute it in the given equation to convert it into a simplified form.

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

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

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

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

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

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

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

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

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

= 1 + sec θ cosec θ = R.H.S.

Hence proved.

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

L.H.S. = $\frac{1+secA}{secA}$

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

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

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

We know that sin²A = (1 – cos²A), we get

= $\frac{1-{\cos }^{2}A}{1-\cos A}$

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

= cos A + 1

L.H.S. = R.H.S.

Hence proved.

(v) $\frac{\cos A-\sin A+1}{\cos A+\sin A-1}$ = cosec A + cot A, using the identity cosec²A = 1+cot²A.

L.H.S. = $\frac{\cos A-\sin A+1}{\cos A+\sin A-1}$

Divide the numerator and denominator by sin A, and we get

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

We know that $\frac{\cos A}{\sin A}$ = cot A and$\frac{1}{\sin A}$ = cosec A

=$\frac{cotA-1+\cos ecA}{cotA+1-\cos ecA}$

=$\frac{cotA-\cos e{c}^{2}A+co{t}^{2}A+\cos ecA}{cotA+1-\cos ecA}$(using cosec²A – cot²A = 1)

=$\frac{(cotA+\cos ecA)-(\cos e{c}^{2}A-co{t}^{2}A)}{cotA+1-\cos ecA}$

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

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

= cosec A + cot A=R.H.

Hence proved.

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

L.H.S.=$\sqrt{\frac{1+\sin A}{1-\sin A}}$

First, divide the numerator and denominator of L.H.S. by cos A,

We know that $\frac{1}{\cos A}$= sec A and $\frac{\sin A}{\cos A}$ = tan A, and it becomes,

= $\sqrt{\frac{secA+\tan A}{secA-\tan A}}$

Now using rationalisation, we get

=$\frac{secA+\tan A}{1}$

= sec A + tan A = R.H.S.

Hence proved.

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

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

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

We know that sin²θ = 1-cos²θ

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

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

= tan θ = R.H.S.

Hence proved.

(viii) (sin A + cosec A)² + (cos A + sec A)² = 7+tan²A+cot²A

L.H.S. = (sin A + cosec A)² + (cos A + sec A)

          = (sin²A + cosec²A + 2 sin A cosec A) + (cos²A + sec²A + 2 cos A sec A)

= (sin²A + cos²A) + 2 sin A($\frac{1}{\sin A}$) + 2 cos A($\frac{1}{\cos A}$) + 1 + tan²A + 1 + cot²A

= 1 + 2 + 2 + 2 + tan²A + cot²A

= 7+tan²A+cot²A = R.H.S.

Hence proved.

(ix) (cosec A – sin A)(sec A – cos A) = $\frac{1}{\tan A+cotA}$.

L.H.S. = (cosec A – sin A)(sec A – cos A)

Now, substitute the inverse and equivalent trigonometric ratio forms.

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

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

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

= cos A sin A

Now, simplify the R.H.S.

R.H.S. = $\frac{1}{\tan A+cotA}$

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

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

= cos A sin A

L.H.S. = R.H.S.

Hence proved.

(x) $\frac{1+{\tan }^{2}A}{1+co{t}^{2}A}$ = ${\left(\frac{1-\tan A}{1-cotA}\right)}^2$ = tan²A

L.H.S. = $\frac{1+{\tan }^{2}A}{1+co{t}^{2}A}$

Since the cot function is the inverse of the tan function,

= $\frac{1+{\tan }^{2}A}{1+{\displaystyle \frac{1}{{\tan }^{2}A}}}$

=$\frac{1+{\tan }^{2}A}{{\displaystyle \frac{1+{\tan }^{2}A}{{\tan }^{2}A}}}$

Now cancel the 1+tan²A terms, and we get

= tan²A

$\frac{1+{\tan }^{2}A}{1+co{t}^{2}A}$ = tan²A

Similarly,

${\left(\frac{1-\tan A}{1-cotA}\right)}^2$ = tan²A

Hence proved.