prove that 1_sin A/1+sinA= (secA-TanA)
Answers
Answered by
2
there will be square of (sec A- tan A).
Attachments:
honey67:
you are right there is square sry
i am sorry
but thanks for helping me
Answered by
1
(1-sinA)/(1+sinA)
Dividing numerator and denominator by cosA
[(1-sinA)cosA]/[(1+sinA)cosA]
=(secA-tanA)/(secA+tanA)_______(1)
∵1+tan²A = sec²A
⇒sec²A - tan²A = 1
Rationalising the equation (1).
[(secA-tanA)(secA-tanA)]/[(secA+tanA)(secA-tanA)]
=(secA-tanA)²/(sec²A-tan²A)
=(secA-tanA)²
=RHS
Dividing numerator and denominator by cosA
[(1-sinA)cosA]/[(1+sinA)cosA]
=(secA-tanA)/(secA+tanA)_______(1)
∵1+tan²A = sec²A
⇒sec²A - tan²A = 1
Rationalising the equation (1).
[(secA-tanA)(secA-tanA)]/[(secA+tanA)(secA-tanA)]
=(secA-tanA)²/(sec²A-tan²A)
=(secA-tanA)²
=RHS
Similar questions