Math, asked by prarthanatank999, 5 months ago

Plz answer this
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Answers

Answered by MissZiddi

LHS=secA+tanA RHS= cos/(1-sinA)

Step-by-step explanation:

Taking, LHS=

as, {secA=1/cosA} and {tanA=sinA/cosA}

=(1/cosA)+(sinA/cosA)

=(1+sinA)/cosA

on Rationalising denominator;

=[(1+sinA)/cosA]×[cosA/cosA]

=[(1+sinA)cosA]÷cos²A

as, sin²A+cos²A=1

=[(1+sinA)cosA]÷(1-sin²A)

using identity, a²–b²=(a+b)(a-b)

=[(1+sinA)cosA]÷[(1+sinA)(1-sinA)]

=cos/1-sinA

Hence, proved

Answered by Anonymous

Sec Θ + tan Θ = cos Θ / 1 - sin Θ

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Answered by Kshitij Duggal ®

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