Question

answer please
i don't known this
please help me​

Answer 1

Step-by-step explanation:

it's cos x not cos ²x in denominator

now ,take LCM

and identity used: sin²a +cos²a = 1

see attachment for your answer

and plz like the answer

Answer 2

Correct Question :-

Prove :- (1+sinA)/cosA + cosA/(1+sinA) = 2secA

Solution :-

Solving LHS,

→ (1+sinA)/cosA + cosA/(1+sinA)

Taking LCM,

→ [(1+sinA)² + cos²A] / cosA.(1+sinA)

using (a + b)² = a² + b² + 2ab in Numerator Now,

→ [1+sin²A+2sinA + cos²A] / cosA.(1+sinA)

Putting sin²A +cos²A = 1 in Numerator Now,

→ (1+1+2sinA)/cosA(1+sinA)

→ (2 + 2sinA) / cosA(1+sinA)

Taking 2 common From Numerator,

→ 2(1+sinA)/cosA(1+sinA)

cancel (1 + sinA) from N & D,

→ (2/cosA)

using (1/cosA) = secA now,

→ 2secA = RHS (Proved).