If secx=p+4p1, then secx+tanx

Given secx=p+4p1 sec2x=p2+16p21+2(p)(4p1)1+tan2x=p2+16p21+21tan2x=p2+16p21+21−1tan2x=p2+16p21−21tan2x=p2+16p21−2(p)(4p1)tan2x=(p−4p1)2tanx=±(p−4p1)secx+tanx ⟹p+4p1 ±(p−4p1)⟹2p(or)2p1

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If x=p+ 1{4p}, then x+ x

Question

If $sec x = p + \frac{1}{4 p}$ , then $sec x + tan x$

2 p or 1/2 p

p or 1/2 p

2 p or 1/ p

None of these

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sec x = p + \frac{1}{4 p}

, then

sec x + tan x

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Text solutionVerified

Given

sec x = p + \frac{1}{4 p}

sec^{2} x = p^{2} + \frac{1}{16 p ^{2}} + 2 (p) (\frac{1}{4 p})

1 + tan^{2} x = p^{2} + \frac{1}{16 p ^{2}} + \frac{1}{2}

tan^{2} x = p^{2} + \frac{1}{16 p ^{2}} + \frac{1}{2} - 1

tan^{2} x = p^{2} + \frac{1}{16 p ^{2}} - \frac{1}{2}

tan^{2} x = p^{2} + \frac{1}{16 p ^{2}} - 2 (p) (\frac{1}{4 p})

tan^{2} x = (p - \frac{1}{4 p})^{2}

tan x = \pm (p - \frac{1}{4 p})

sec x + tan x

⟹ p + \frac{1}{4 p} \pm (p - \frac{1}{4 p})

⟹ 2 p (or) \frac{1}{2 p}

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Question Text	If $sec x = p + \frac{1}{4 p}$ , then $sec x + tan x$
Answer Type	Text solution:1
Upvotes	150

If secx=p+4p1​, then secx+tanx

Text solutionVerified

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If $sec x = p + \frac{1}{4 p}$ , then $sec x + tan x$