sec'A (1 - sin'A) - 2tan' = 1
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Answer:
Taking LHS [we know 1+tan
2
θ−sec
2
θ]
⇒sec
4
A(1−sin
4
A)−2tan
2
A
⇒
cos
4
A
1
(1−sin
4
A)−2tan
2
A
=sec
4
A−tan
4
A−2tan
2
A
⇒(1+tan
2
A)
2
−tan
4
A−2tan
2
A
⇒1+2tan
2
A+tan
4
A−tan
4
A−2tan
2
A
⇒1 RHS proved
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