Answer:
I believe the answer is d
1)
LHS = cot(a/2) - tan(a/2)
= (1 - tan^2(a/2))/tan(a/2)
= (2-sec^2(a/2))/tan(a/2)
= 2cot(a/2) - cosec(a/2)sec(a/2)
= 2(1+cos(a))/sin(a) - 1/(cos(a/2)sin(a/2))
= 2 (1+cos(a))/sin(a) - 2/sin(a)) (product to sums)
= 2[(1+cos(a) -1)/sin(a)]
=2cot a
= RHS
2.
LHS = cot(b/2) + tan(b/2)
= [1 + tan^2(b/2)]/tan(b/2)
= sec^2(b/2)/tan(b/2)
= 1/sin(b/2)cos(b/2)
using product to sums
= 2/sin(b)
= 2cosec(b)
= RHS
Answer:
30, 30, 120
Step-by-step explanation:
Factor out the cos<span>θ:
</span>cosθ (2sin<span>θ + sqrt3) = 0
</span>Therefore, the only ways this can happen are if either cosθ = 0 or if (2sin<span>θ + sqrt3) = 0
</span>The first case, cosθ = 0 only at θ <span>= pi/2, 3pi/2.
</span>The second case, <span>(2sin<span>θ + sqrt3) = 0 simplifies to:
</span></span>sin<span>θ = (-sqrt3)/2
</span><span><span>θ = 4pi/3, 5pi/3
</span></span><span><span>Therefore the answer is A.
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