Solve: if you can
Prove that: sec^2 theta - ((sin^2 theta - 2 sin^4 theta) /(2 cos^4 theta - cos^2 theta)) = 1
Observe the question carefully
ursamajor:
I am sorry but this topic has not been explained to me yet. I am in class 10 and I have done only first four chapters. You can ask anything from them.
Answers
Answered by
175
HELLO DEAR,
GIVEN:-
sec² θ - {(sin²θ - 2 sin⁴θ ) /(2 cos⁴θ - cos²θ )}
=> sec²θ - {sin²θ(1 - 2sin²θ)/cos²θ(2cos²θ - 1)}
=> sec²θ - [tan²θ × {1 - 2(1 - cos²θ)}/(2cos²θ - 1)]
=> sec²θ - [tan²θ × {1 - 2 + 2cos²θ}/{2cos²θ - 1}]
=> sec²θ - [tan²θ × (2cos²θ - 1)/(2cos²θ - 1)]
=> sec²θ - tan²θ
=> 1 [as, 1 + tan²θ = 1]
I HOPE IT'S HELP YOU DEAR,
THANKS
GIVEN:-
sec² θ - {(sin²θ - 2 sin⁴θ ) /(2 cos⁴θ - cos²θ )}
=> sec²θ - {sin²θ(1 - 2sin²θ)/cos²θ(2cos²θ - 1)}
=> sec²θ - [tan²θ × {1 - 2(1 - cos²θ)}/(2cos²θ - 1)]
=> sec²θ - [tan²θ × {1 - 2 + 2cos²θ}/{2cos²θ - 1}]
=> sec²θ - [tan²θ × (2cos²θ - 1)/(2cos²θ - 1)]
=> sec²θ - tan²θ
=> 1 [as, 1 + tan²θ = 1]
I HOPE IT'S HELP YOU DEAR,
THANKS
Answered by
208
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