Prove that : sec* 4A(1-sin*4 A)-2 tan*2 A=1
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Answer:
Consider the L.H.S,
=sec
4
A(1−sin
4
A)−2tan
2
A
=sec
4
A−sec
4
Asin
4
A−2tan
2
A
=sec
4
A−
cos
4
A
sin
4
A
−2tan
2
A
=sec
4
A−tan
4
A−2tan
2
A
=[(sec
2
A)
2
−(tan
2
A)
2
]−2tan
2
A
=[(sec
2
A)−(tan
2
A)][(sec
2
A)+(tan
2
A)]−2tan
2
A
We know that
sec
2
A−tan
2
A=1
Therefore,
=sec
2
A+tan
2
A−2tan
2
A
=sec
2
A−tan
2
A
=1
Hence, proved
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