Question

$\frac{si{n}^{2}A-si{n}^{2}B}{sinAcosA-sinBcosB}$ = a when A = ${20}^{\circ }$ and B = ${25}^{\circ }$.Find the value of $\frac{1}{a^2}$.

__

Answer 1

We are given A = ${20}^{\circ }$ and B=${25}^{\circ }$. We will use this condition after simplifying the given expression. We also note that A+B=${45}^{\circ }$.

$si{n}^{2}A$ - $si{n}^{2}B$ can be simplified as sin (A+B) sin (A-B). [We have sinAcosA and sinBcosB. After simplifying there are some chances of sin (A+B) or sin (A-B) getting cancelled from numerator).

$\Rightarrow$ $\frac{si{n}^{2}A-si{n}^{2}B}{sinAcosA-sinBcosB}$ = $\frac{sin(A+B)sin(A-B)}{\frac{sin2A-sin2B}{2}}$

[Now we can apply transformation formula]

= $\frac{2sin(A+B)sin(A-B)}{2sin(A-B)cos(A+B)}$

= tan(A+B)

A = ${20}^{\circ }$, B = ${25}^{\circ }$

$\Rightarrow$ tan(A+B) = 1 = a

$\Rightarrow$ $\frac{1}{a^2}$ = 1

Key steps/concepts: (1) $si{n}^{2}A$ - $si{n}^{2}B$ = sin(A+B) $\times$ sin(A-B)

(2) sinAcosA = $\frac{1}{2}$ sin2A

(3) sinA - sinB = 2sin$\frac{(A-B)}{2}$cos$\frac{(A+B)}{2}$