If tan (θ/2)= operatornamecosec θ- sin θ, then -

Question

If tan (θ/2)= operatornamecosec θ- sin θ, then - (A)  sin 2 (θ/2)=2 sin 2 18° (B)  cos 2 θ+2 cos θ+1=0 (C)  sin 2 (θ/2)=4 sin 2 18° (D)  cos 2 θ+2 cos θ-1=0

Tardigrade Admin · Accepted Answer

Correct answer is (a) sin 2 (θ/2)=2 sin 2 18°Correct answer is (d) cos 2 θ+2 cos θ-1=0

Q. If $tan \frac{θ}{2} = cosec θ - sin θ$ , then -

$sin^{2} \frac{θ}{2} = 2 sin^{2} 1 8^{\circ}$

$cos 2 θ + 2 cos θ + 1 = 0$

$sin^{2} \frac{θ}{2} = 4 sin^{2} 1 8^{\circ}$

$cos 2 θ + 2 cos θ - 1 = 0$

Correct answer is (a) $sin^{2} \frac{θ}{2} = 2 sin^{2} 1 8^{\circ}$ Correct answer is (d) $cos 2 θ + 2 cos θ - 1 = 0$

Q. If tan2θ​=cosecθ−sinθ, then -

sin22θ​=2sin218∘

cos2θ+2cosθ+1=0

sin22θ​=4sin218∘

cos2θ+2cosθ−1=0