Question

Prove that √((secA - tanA)/(secA + tanA)) = cosA/(1+sinA)

No absurd answers please.

Answer 1

Answer:

Step-by-step explanation:

multiply and divide the question by root over secA-tanA we get

√((secA-tanA)²/(sec²A-tan²A)) and we know that the denominator is a trigonometric identity ; so denominator equals 1 and the expression becomes

(secA-tanA)   [the root gets cancelled with square]

secA=1/cosA   and tanA=sinA/cosA

plugging in we get

(1-sinA)/cosA--------(1)

now we know sin²A+cos²A=1 so

cos²A=1-sin²A further we get

cos²A=(1+sinA)(1-sinA)    so 1-sinA will be

cos²A/(1+sinA) now plugging the 1-sinA value in (1) we get

cos²A/(1+sinA)*1/cosA as we observe one cosA in the denominator gets cancelled with the cosA in the numerator leaving a single cosA in the numerator so the final value will be

cosA/(1+sinA) hence L.H.S=R.H.S

hence proved

I REALLY HOPE THIS HELPS YOU