If sin⁡ (θ+α)=a,cos2 ⁡(θ+β)=b, then sin ⁡(α−β)=

A
$a b - (1 - a^{2}) (1 - b^{2})$
B
$a b - \sqrt{(1 - a^{2}) (1 - b^{2})}$
C
$\pm \{a \sqrt{b} - \sqrt{(1 - a^{2}) (1 - b)}\}$
D
$b \sqrt{a} - \sqrt{(1 - a)^{2} (1 - b)}$

Solution:

We have,

$\begin{array}{l} \sin (θ + α) = a and \cos (θ + β) = \pm \sqrt{b} \\ ∴ \cos (θ + α) = \pm \sqrt{1 - a^{2}} and, \sin (θ + β) = \pm \sqrt{1 - b} \end{array}$

Now,

$\begin{array}{l} \sin (α - β) = \sin {(θ + α) - (θ + β)} \\ \Rightarrow \sin (α - β) = \sin (θ + α) \cos (θ + β) - \cos (θ + α) \sin (θ + β) \\ \Rightarrow \sin (α - β) = \pm a \sqrt{b} - \{\pm \sqrt{1 - a^{2}}\} \times {\pm \sqrt{1 - b}} \\ \Rightarrow \sin (α - β) = \pm \{a \sqrt{b} - \sqrt{1 - a^{2}} \sqrt{1 - b}\} \end{array}$

If $\sin (θ + α) = a, \cos^{2} (θ + β) = b,$ then $\sin (α - β) =$

Solution:

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