Question

prove that:-
$\tt{(cosec \;theta + cot \;theta) ^2 = sec \:theta +1 / sec \;theta - 1​}$​

Answer 1

ᴛᴏ ᴘʀᴏᴠᴇ:-

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

ᴘʀᴏᴏꜰ:−

$\tt LHS = {(cosec \theta + cot \theta)}^{2}$

$\tt= { \bigg(\dfrac{1}{sin\theta }+ \dfrac{cos\theta}{sin \theta} \bigg)}^{2}$

$\tt= { \bigg( \dfrac{1 + cos\theta}{sin \theta} \bigg)}^{2}$

$\tt = { \dfrac{(1 + cos\theta)^{2} }{sin ^{2} \theta}}$

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

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

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

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

$\tt = \dfrac{1 + \dfrac{1}{sec \theta} }{1 - \dfrac{1}{sec \theta} }$

$\tt = \dfrac{ \dfrac{sec \theta + 1}{sec \theta} }{ \dfrac{sec \theta - 1}{sec \theta} }$

$\tt= \dfrac{ sec \theta + 1}{ sec \theta - 1}$

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$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}}$

Answer 2

Answer:

$\tt{(cosec \;theta + cot \;theta) ^2 = sec \:theta +1 / sec \;theta - 1}$