Question

please tell me how to prove this​

Answer 1

Solution: Let's solve L.H.S first,

L.H.S. → tan²A/(tan²A - 1) + csc²∅/(sec²A - csc²∅)

• tanA = sinA/cosA
• cscA = 1/sinA
• secA = 1/cosA

→ (sin²A/cos²A)/(sin²A/cos²A - 1) + (1/sin²A)/(1/cos²A - 1/sin²A)

On taking L.C.M. of denominators

→ (sin²A/cos²A)/(sin²A - cos²A)/cos²A + (1/sin²A)/(sin²A - cos²A)/cos²A sin²A

Changing division sign into multiplication by reciprocal

→ (sin²A/cos²A) × cos²A/(sin² - cos²A) + (1/sin²A) × cos²A sin²A/(sin²A - cos²A)

→ sin²A/(sin² - cos²A) + cos²A/(sin² - cos²A)

Taking denominators as common

→ (sin²A + cos²A)/(sin²A - cos²A)

Using identity, sin²A + cos²A = 1

→ 1/(sin²A - cos²A)

“sin²A + cos²A = 1 → sin²A = 1 - cos²A”

→ 1/(1 - cos²A - cos²A)

→ 1/(1 - 2cos²A) = R.H.S.

Answer 2

Answer:

heres your answer buddy...

take care : )