Question

Please answer Question B2 and B3...

Answer 1

B-2 :

Consider the Identity : cos²θ - sin²θ = cos2θ

★  We know that : cos²θ = 1 - sin²θ

Substitute the value of cos²θ in the identity of cos2θ

$:\implies$  1 - sin²θ - sin²θ = cos2θ

$:\implies$  1 - 2sin²θ = cos2θ

$:\implies$  2sin²θ = 1 - cos2θ

$:\implies \mathsf{sin^2\theta = \left(\dfrac{1 - cos2\theta}{2}\right)}$

Answer : Option (3)

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B-3 :

Given : sinA.sin(A + B)

★  We know that : sin(A + B) = sinA.cosB + cosA.sinB

$:\implies$  (sinA)[sinA.cosB + cosA.sinB]

$:\implies$  sin²A.cosB + sinA.cosA.sinB

The above expression can be written as :

$\mathsf{:\implies sin^2A.cosB + \dfrac{2}{2}(sinA.cosA.sinB)}$

$\mathsf{:\implies sin^2A.cosB + \left(\dfrac{2sinA.cosA}{2}\right)(sinB)}$

★  We know that : sin2A = 2sinA.cosA

$\mathsf{:\implies sin^2A.cosB + \left(\dfrac{sin2A}{2}\right)(sinB)}$

$\mathsf{:\implies sin^2A.cosB + \dfrac{1}{2}sin2A.sinB}$

Answer : Option (3)