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# Prove the following identities, where the angles involved are acute angles for which the expressions are defined : (i) $$(cosec \theta - cot \theta )^2 \ = \ \frac{1 - cos \theta}{1 + cos \theta}$$ (ii) $$\frac{cosA}{1+sinA} \ + \ \frac{1+sinA}{cosA} \ = \ 2 secA$$ (iii) $$\frac{tan \theta}{1 - cot\theta} \ + \ \frac{cot \theta}{1 - tan\theta} \ = \ 1 \ + \ sec \theta cosec \theta$$ [Hint : Write the expression in terms of sin $$\theta$$ and cos $$\theta$$] (iv) $$\frac{1 \ + \ secA}{secA} \ = \ \frac{sin^2A}{1 \ - \ cosA}$$ [Hint : Simplify LHS and RHS separately] (v) $$\frac{cosA \ - \ sinA \ + \ 1}{ cosA \ + \ sinA \ - \ 1} \ = \ cosecA \ + \ cotA$$ , using the identity $$cosec^2A \ = \ 1 \ + \ cot^2A$$ (vi) $$\sqrt{ \frac{1 \ + \ sinA}{1 \ - \ sinA}} \ = \ secA \ + \ tanA$$ (vii) $$\frac{sin\theta - 2sin^3 \theta}{2cos^3 \theta - cos\theta} \ = \ tan \theta$$ (viii) $$(sinA \ + \ cosecA)^2 \ + \ (cosA \ + \ secA)^2 \ = \ 7 \ + \ tan^2A \ + \ cot^2A$$ (ix) $$(cosecA \ - \ sinA)(secA \ - \ cosA) \ = \ \frac{1}{tanA \ + \ cotA}$$ [Hint : Simplify LHS and RHS separately] (x) $$( \frac{1 \ + \ tan^2A}{1 \ + \ cot^2A} ) \ = \ ( \frac{1 \ - \ tanA}{1 \ - \ cotA})^2 \ = \ tan^2A$$

(i) L.H.S. = $$(cosec \theta - cot \theta )^2 \$$
$$= \ ( \frac{1}{sin\theta} \ - \ \frac{cos \theta}{ sin \theta} )^2$$
$$= \ ( \frac{1 \ - \ cos \theta }{ sin \theta } )^2$$

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

$$= \ \frac{(1 \ - \ cos\theta )^2}{(1 \ - \ cos\theta)(1 \ + \ cos\theta)} \$$
$$= \ \frac{1 \ - \ cos\theta}{1 \ + \ cos\theta}$$ = R.H.S.
[ $$\because A^2 \ - \ B^2 \ = \ (A+B)(A-B)$$ ]

(ii) L.H.S. = $$\frac{cosA}{1+sinA} \ + \ \frac{1+sinA}{cosA}$$
$$= \ \frac{cos^2A \ + \ (1 \ + \ sinA)^2}{cosA(1 \ + \ sinA)}$$

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

$$= \ \frac{1 \ + \ 1 \ + \ 2sinA}{cosA(1 \ + \ sinA)}$$
[ $$\because sin^2A \ + \ cos^2A \ = \ 1$$ ]

$$= \ \frac{2 \ + \ 2sinA}{cosA(1 \ + \ sinA)} \$$
$$= \ \frac{2(1 \ + \ sinA)}{cosA(1 \ + \ sinA)}$$

$$= \ \frac{2}{cosA} \ = \ 2secA \ =$$ R.H.S.

(iii) L.H.S. = $$\frac{tan \theta}{1 - cot\theta} \ + \ \frac{cot \theta}{1 - tan\theta}$$
$$= \frac{ \frac{sin \theta}{cos\theta}}{ 1 \ - \ \frac{cos\theta}{sin\theta}} \ + \ \frac{ \frac{cos\theta}{sin\theta}}{1 \ - \ \frac{sin\theta}{cos\theta}}$$

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

$$= \ \frac{sin^3 \theta \ - \ cos^3 \theta}{ cos\theta sin\theta( sin\theta \ - \ cos\theta)}$$
$$= \ \frac{(sin\theta \ - \ cos\theta)( sin^2 \theta \ + \ sin\theta cos\theta \ + \ cos^2 \theta)}{( cos\theta sin\theta( sin\theta \ - \ cos\theta)}$$ [ $$\because a^3 \ - \ b^3 \ = \ (a-b)(a^2 \ + \ ab \ + \ b^2)$$ ]

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

$$= \ 1 \ + \ sec \theta cosec \theta$$ = R.H.S.

(iv) L.H.S. $$= \ \frac{1 \ + \ secA}{secA} \$$
$$= \ 1 \ + \ \frac{1}{secA} \ = \ 1 \ + \ cosA$$
$$= \ \frac{(1 \ - \ cosA )(1 \ + \ cosA)}{1 \ - \ cosA}$$

$$= \ \frac{1 \ - \ cos^2A}{1 \ - \ cosA}$$
[ $$\because A^2 \ - \ B^2 \ = \ (A+B)(A-B)$$ ]

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

= R.H.S.

(v) L.H.S. = $$\frac{cosA \ - \ sinA \ + \ 1}{ cosA \ + \ sinA \ - \ 1}$$
$$= \ \frac{ \frac{ cosA \ - \ sinA \ + \ 1}{sinA}}{ \frac{cosA \ + \ sinA \ - \ 1}{sinA}}$$

$$= \ \frac{cotA \ - \ 1 \ + \ cosecA}{cosA \ + \ 1 \ - \ cosecA}$$
[ $$\because 1 \ + \ cot^2A \ = \ cosec^2A$$ ]

$$= \ \frac{cotA \ + \ cosecA \ - \ (cosec^2A \ - \ cot^2A)}{cotA \ - \ cosecA \ + \ 1}$$
$$= \ \frac{cotA \ + \ cosecA \ - \ (cosecA \ + \ cotA)(cosecA \ - \ cotA)}{cotA \ - \ cosecA \ + \ 1}$$
[ $$\because A^2 \ - \ B^2 \ = \ (A+B)(A-B)$$ ]

Taking common(cosecA + cotA)

$$= \ \frac{(cosecA \ + \ cotA)(1 \ - \ cosecA \ + \ cot)}{(cotA \ - \ cosecA \ + \ 1 )}$$
$$= \ cosec A \ + \ cot A$$ = R.H.S.

(vi) L.H.S. = $$\sqrt{ \frac{1 \ + \ sinA}{1 \ - \ sinA}}$$
$$= \ \sqrt{ \frac{1+sinA}{1-sinA} × \frac{1+sinA}{1+sinA}}$$
[Multiplying and dividing by $$\sqrt{1+sinA}$$ ]

$$= \ \sqrt{ \frac{(1+sinA)^2}{1-sin^2A}} \ = \ \sqrt{ \frac{(1+sinA)^2}{cos^2A}}$$
[ $$\because sin^2A \ + \ cos^2A \ = \ 1$$ ]

$$= \ \frac{1+sinA}{cosA}$$
$$= \ \frac{1}{cosA} \ + \ \frac{sinA}{cosA}$$

$$= \ secA \ + \ tanA$$
[ $$\because tanA \ = \ \frac{sinA}{cosA}$$ and $$secA \ = \ \frac{1}{cosA}$$ ]

= R.H.S.

(vii) L.H.S. = $$\frac{sin\theta - 2sin^3 \theta}{2cos^3 \theta - cos\theta}$$
$$= \ \frac{ sin \theta(1-2sin^2 \theta)}{cos \theta (2cos^2 \theta -1)}$$

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

$$= \ tan\theta$$

= R.H.S.

(viii) L.H.S. =$$(sinA \ + \ cosecA)^2 \ + \ (cosA \ + \ secA)^2$$
$$= \ (sin^2A \ + \ cosec^2A \ + \ 2sinAcosecA) \ + \ (cos^2A \ + \ sec^2A \ + \ 2cosAsecA)$$

$$= \ (sin^2A \ + \ cosec^2A \ + \ 2 ) \ + \ (cos^2A \ + \ sec^2A \ + \ 2 )$$
$$= \ 1+ cosec^2A \ + \ sec^2A \ + \ 4$$ [ $$\because sin^2A \ + \ cos^2A \ = \ 1$$ ]

$$= \ 1 \ + \ cot^2A \ + \ 1 \ + \ tan^2A \ + \ 5$$ [ $$\because sec^2A \ = \ 1 \ + \ tan^2A$$ and $$cosec^2A \ = \ 1 \ + \ cot^2A$$ ]

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

(ix) L.H.S. = $$(cosecA \ - \ sinA)(secA \ - \ cosA)$$
$$= ( \frac{1}{sinA} \ - \ sinA)( \frac{1}{cosA} \ - \ cosA)$$

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

$$= \ sinA cosA$$
$$= \ \frac{sinAcosA}{sin^2A \ + \ cos^2A}$$
[ $$\because sin^2A \ + \ cos^2A \ = \ 1$$ ]

Dividing Numerator and Denominator by $$sinA cosA$$, we get

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

$$= \ \frac{1}{tanA \ + \ cotA}$$ = R.H.S

(x) L.H.S. = $$( \frac{1 \ + \ tan^2A}{1 \ + \ cot^2A} )$$
$$= \ \frac{sec^2A}{cosec^2A}$$

$$= \ \frac{1}{cos^2A} \ × \ sin^2A \ = \ tan^2A$$ [ $$\because secA \ = \ \frac{1}{cosA}$$ and $$cosecA \ = \ \frac{1}{sinA}$$ ]

R.H.S. = $$\frac{1 \ - \ tanA}{1 \ - \ cotA})^2$$
$$= \ ( \frac{1 \ - \ tanA}{ 1 \ - \ \frac{1}{tanA}})^2$$

$$= \ ( \frac{1 \ - \ tanA}{ \frac{tanA \ - \ 1}{tanA}})^2$$ $$= \ tan^2A$$