Question

(cosec-cos)^2 = 1-cos/1+cos Prove it

Answer 1

Correct Question

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

Solution

From LHS,

$\tt = (cosec \theta - cot \theta) {}^{2} \\ \\ \tt = cosec {}^{2} \theta \: + cot {}^{2} \theta - 2cosec \theta \: cot \theta \\ \\ \tt = \frac{1}{sin {}^{2} \theta } + \frac{cos {}^{2} \theta}{sin {}^{2} \theta } - 2 \times \frac{1}{sin \theta} \times \frac{cos \theta}{sin \theta} \\ \\ \tt = \frac{1 + cos {}^{2} theta }{sin {}^{2} \theta} - \frac{2cos \theta}{sin {}^{2} \theta } \\ \\ \tt = \frac{ 1 + cos {}^{2} \theta - 2cos \theta}{sin {}^{2} \theta} \\ \\ \tt = \frac{(1 - cos \theta) {}^{2} }{sin {}^{2} \theta} \\ \\ \tt = \frac{(1 - cos \theta)(1 - cos \theta}{1 - cos {}^{2} \theta } \\ \\ \tt = \frac{(1 - cos \theta)(1 - cos \theta}{(1 + cos \theta)(1 - cos \theta} \\ \\ \tt = \frac{1 - cos \theta}{1 + cos \theta} = RHS$

$\large \implies {\boxed {\tt \blue {Hence \: Proved }}}$

Used identities

$\tt \implies [Used \: in \: 2nd \: step] \\ (a + b) {}^{2} = a {}^{2} + b {}^{2} + 2ab \\ \\ \tt \implies [used \: in \: 3rd \: step] \\ cosec {}^{2} \theta = \frac{1}{sin {}^{2} \theta } \\ cot {}^{2} \theta = \frac{sin {}^{2} \theta}{cos {}^{2} \theta}$

Answer 2

${\huge{\bold{\blue{\boxed{\bf{Correct\: Question}}}}}}$

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

${\huge{\bold{\pink{\bigstar{\boxed{\bf{Solution}}}}}}}$

${\bold{\green{\boxed{\bf{LHS}}}}}$

$\implies (cosec \theta - cot \theta) {}^{2}$

• $(x+y)^2 = x^2 + y^2 + 2xy$

$\implies cosec {}^{2} \theta \: + cot {}^{2} \theta - 2cosec \theta \: cot \theta$

• $cosec^2\theta = \dfrac{1}{sin^2\theta}$

• $cot^2\theta = \dfrac{sin^2\theta}{cos^2\theta}$

$\implies \frac{1}{sin {}^{2} \theta } + \dfrac{cos {}^{2} \theta}{sin {}^{2} \theta } - 2 \times \dfrac{1}{sin \theta} \times \dfrac{cos \theta}{sin \theta}$

$\implies \dfrac{1 + cos {}^{2} \theta }{sin {}^{2} \theta} - \dfrac{2cos \theta}{sin {}^{2} \theta }$

$\implies \dfrac{ 1 + cos {}^{2} \theta - 2cos \theta}{sin {}^{2} \theta}$

$\implies \dfrac{(1 - cos \theta) {}^{2} }{sin {}^{2} \theta}$

$\implies\dfrac{(1 - cos \theta)(1 - cos \theta}{1 - cos {}^{2} \theta }$

$\implies \dfrac{(1 - cos \theta)(1 - cos \theta}{(1 + cos \theta)(1 - cos \theta}$

$\implies\dfrac{1 - cos \theta}{1 + cos \theta}$

${\bold{\green{\boxed{\bf{RHS}}}}}$

$\implies\dfrac{1 - cos \theta}{1 + cos \theta}$

${\bold{\blue{\boxed{\bf{COMPARE}}}}}$

$\implies\dfrac{1 - cos \theta}{1 + cos \theta}$ = $\dfrac{1 - cos \theta}{1 + cos \theta}$

${\bold{\blue{\boxed{\bf{HencE\: Pr0ved}}}}}$

...........hence proved

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