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An Intiative by Sri Chaitanya
a
122
b
7−1
c
22−1
d
17
answer is D.
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Detailed Solution
14sin−1638=θsin4θ=6382tan2θ1+tan22θ=638631+tan22θ=16tan2θ63tan22θ−16tan2θ+63=0Use the quadratic formula for It gives 2tanθ1−tan2θ=963 or 7637tan2θ−1+263tanθ=0tanθ=−263±252+19614=−263±44814=−67±8714=17