Question

Plzz help me with this trig question...

plz try showing me how to prove from both the sides ...LHS to RHS and RHS to LHS​

Answer 1

Explanation:

(sec(8A) − 1) / (sec(4A) − 1)

Multiply by the reciprocal:

(sec(8A) − 1) / (sec(4A) − 1) × (sec(4A) + 1) / (sec(4A) + 1)

(sec(8A) − 1) (sec(4A) + 1) / (sec²(4A) − 1)

(sec(8A) − 1) (sec(4A) + 1) / tan²(4A)

According to double angle formula, sec(2x) = sec² x / (2 − sec² x). So we can rewrite the first term as:

sec(8A) − 1

sec² (4A) / (2 − sec² (4A)) − 1

sec² (4A) / (2 − sec² (4A)) − (2 − sec² (4A)) / (2 − sec² (4A))

(sec² (4A) − 2 + sec² (4A)) / (2 − sec² (4A))

(2 sec² (4A) − 2) / (2 − sec² (4A))

2 (sec² (4A) − 1) / (2 − sec² (4A))

2 tan² (4A) / (2 − sec² (4A))

Substituting:

2 (sec(4A) + 1) / (2 − sec² (4A))

Rearranging:

2 (sec(4A) + 1) / (1 + 1 − sec² (4A))

2 (sec(4A) + 1) / (1 − tan² (4A))

From half angle formula, we know tan(x/2) = tan x / (1 + sec x). So we can rewrite the numerator as:

sec(4A) + 1

tan(4A) / tan(2A)

Substituting:

2 tan(4A) / (tan(2A) (1 − tan² (4A)))

Using double angle formula:

tan(8A) / tan(2A)