Question

prove that (codec theta + cot theta) ^2 = sec theta +1 / sec theta - 1​

Answer 1

Step-by-step explanation:

$\sf \underline{To\:Prove:-}$

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

Formulas used:-

$\sf cosec\theta = \dfrac{1}{sin\theta}$

$\sf cot\theta = \dfrac{cos\theta}{sin\theta}$

$\sf \underline{Proof:-}$

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

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

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

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

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

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

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

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

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

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

$\sf = \dfrac{ sec \theta + 1}{ sec \theta - 1} =RHS$

$\sf LHS = RHS$

$\sf \underline{Hence\: Proved}$

Answer 2

Answer:

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