If secx+tanx=p then prove that sinx=p2+1p2−1

secx+tanx=psecx−tanx=1/p2secx=p+p1=pp2+1secx=2pp2+1cosx=p2+12pAC2=AB2+BC2(p2+1) 2=(2p) 2+(AB) 2p4+1+2p2−4p2=AB2AB=p2−1∴sinx=p2+1p2−1

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If x+ x=p then prove that x= {{p}^{2}-1}{{p}^{2}+1}

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If $sec x + tan x = p$ then prove that $sin x = \frac{p ^{2} - 1}{p ^{2} + 1}$

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Owen Discussed

sec x + tan x = p

then prove that

sin x = \frac{p ^{2} - 1}{p ^{2} + 1}

12 mins ago

Discuss this question LIVE

12 mins ago

sec x + tan x = p

sec x - tan x = 1/ p

2 sec x = p + \frac{1}{p} = \frac{p ^{2} + 1}{p}

sec x = \frac{p ^{2} + 1}{2 p}

cos x = \frac{2 p}{p ^{2} + 1}

A C^{2} = A B^{2} + BC^{2}

(p^{2} + 1)^{2} = (2 p)^{2} + (A B)^{2}

p^{4} + 1 + 2 p^{2} - 4 p^{2} = A B^{2}

A B = p^{2} - 1

∴ sin x = \frac{p ^{2} - 1}{p ^{2} + 1}

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Question Text	If $sec x + tan x = p$ then prove that $sin x = \frac{p ^{2} - 1}{p ^{2} + 1}$
Answer Type	Text solution:1
Upvotes	150