Question

if secA-tanA=K then prove that sinA=1-k/1+k square

Answer 1

Answer:

Here is your answer...

K = sec A -tan A

R. H. S = 1- k²/1+k²

= 1-(sec A - tan A) ²/1+ (sec A -TanA ) ²

= 1- (sec A²+tanA²-2tanA Sec A) /1+(Sec A²+

tanA² -2 tan A Sec A)

(Since ( a-b) ² = a²+b²-2ab)

= 1- (tan A ² -1 +tan A²-2 tan A sec A /1+( sec A ²

1+tan A² -2 TanA sec A)

=2tan A² -2tan A sec A / 2 sec A² -2 tan A sec A

= 2tan A (Sec A -tan A) /2 sec A (sec A - tan A)

= Tan A / sec A

= sin A/ cos A/1/ cos A

= sin A/cos A × cos A/1

= sin A /1

= sin A = R. H. S

Hence proved....

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