Question

if secθ + tan θ = m, show that (m²-1)/(m²+1) = sin θ​

Answer 1

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Answer 2

Given that,

secθ + tan θ = m

$\sf ( {m}^{2} - 1) = (sec \theta + tan \theta)^{2} - 1$

$\sf = sec ^{2} \theta + tan^{2} \theta + 2sec \theta \: tan \theta - 1$

$\sf =( sec ^{2} \theta - 1) + tan^{2} \theta + 2sec \theta \: tan \theta$

$\sf =tan^{2} \theta+ tan^{2} \theta + 2sec \theta \: tan \theta$

$\sf =2tan^{2} \theta + 2sec \theta \: tan \theta$

$\sf =2tan\theta (tan \theta + \: sec \theta) \: \: ...(1)$

similarly,.

$\sf ( {m}^{2} + 1) = (sec \theta + tan \theta)^{2} + 1$

$\sf = sec ^{2} \theta + tan^{2} \theta + 2sec \theta \: tan \theta + 1$

$\sf =(tan^{2} \theta + 1) + sec^{2} \theta + 2sec \theta \: tan \theta$

$\sf =sec^{2} \theta+ sec^{2} \theta + 2sec \theta \: tan \theta$

$\sf =2sec {}^{2} \theta + 2sec \theta \: tan \theta$

$\sf =2sec\theta (tan \theta + \: sec \theta) \: \: ...(2)$

From equation (1) and (2)

$\sf \dfrac{ {m}^{2} - 1}{ {m}^{2} + 1} = \dfrac{2tan\theta (tan \theta + \: sec \theta)}{2sec\theta (tan \theta + \: sec \theta)}$

$\sf = \dfrac{tan\theta }{sec\theta }$

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

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

$\sf = sin \theta$

Thus,

$\sf \dfrac{ {m}^{2} - 1}{ {m}^{2} + 1} = sin\theta$