Question

solve the question✔✔✔✔



tan²theta / (sec theta-1) ²= 1+cos theta /1-cos theta ​

Answer 1

Answer:

o bhai ye question ha???

pls don't report pls follow

Answer 2

Question: -

$\bf \implies \: \dfrac{ { \tan( \theta) }^{2} }{ \sec( \theta - 1)^{2} } = \dfrac{1 + \cos( \theta) }{1 - \cos( \theta) }$

LHS:-

$\implies \sf \: \dfrac{ \dfrac{ {sin}^{2} \theta }{ {cos}^{2} \theta } }{ \big(\dfrac{1}{cos \theta} - 1 \big)^{2} } = \dfrac{ \dfrac{ {sin}^{2} \theta }{ {cos}^{2} \theta } }{ \dfrac{(1 - cos \theta)^{2} }{cos ^{2} } \theta}$

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

RHS:-

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

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

LHS = RHS, hence proved!!