Question

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Answer 1

Trigonometry

• We will be using the following identities:

• $\boxed{\begin{minipage}{20 em} \sf \displaystyle \bullet \sin(A+B)=\sin A\cos B+\cos A\sin B \\\\\\ \bullet \cos(A-B)=\cos A\cos B+\sin A\sin B \\\\\\ \bullet \sin 2\theta = 2\sin \theta \cos\theta \\\\\\ \bullet \sin C+\sin D=2\sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right) \end{minipage}}$

• We can start by looking at what we have and what we want.

• $\sf\sin A+\sin B=a \quad\textsf{----- (1)}\\\\ \cos A+\cos B=b \quad \textsf{----- (2)}$

On the RHS, we see two specific terms:

• $\sf ab$ in the numerator and $\sf a^2+b^2$ in the denominator. Let's find out the values:

• $\begin{aligned}\sf\right arrow ab &= \sf(\sin A+\sin B)(\cos A+\cos B)\\\\ &= \sf \sin A\cos A+\sin A\cos B+\cos A\sin B+\sin B\cos B \\\\ \therefore \sf ab&= \sf \sin (A+B)+\sin A\cos A+\sin B\cos B \end{aligned}$

• $\begin{aligned}\sf\rightarrow a^2+b^2 &= \sf(\sin A+\sin B)^2+(\cos A+\cos B)^2\\\\ &= \sf \sin^2 A+\sin^2B+2\sin A\sin B+\cos^2A+\cos^2B+2\cos A\cos B \\\\ &=\sf (\sin^2 A+\cos^2A)+(\sin^2B+\cos^2B)+2(\cos A\cos B+\sin A\sin B) \\\\ &=\sf 1+1+2\cos(A-B) \\\\ \therefore \sf a^2+b^2&= \sf 2(1+\cos(A-B))\end{aligned}$

Now let's take the RHS:-

$\sf\displaystyle \mathbb{RHS}\\\\\\ = \frac{2ab}{a^2+b^2} \\\\\\ = \frac{2(\sin (A+B)+\sin A\cos A+\sin B\cos B)}{2(1+\cos(A-B))}\\\\\\ = \frac{\sin(A+B)+\frac{1}{2}(2\sin A\cosA + 2\sin B\cos B)}{1+\cos(A-B)}\\\\\\ = \frac{\sin(A+B)+\frac{1}{2}(\sin 2A+\sin 2B)}{1+\cos(A-B)}\\\\\\ = \frac{\sin(A+B)+\frac{1}{2}\times 2 \sin\left(\frac{2A+2B}{2}\right)\cos\left(\frac{2A-2B}{2}\right)}{1+\cos(A-B)}$

$\sf\displaystyle = \frac{\sin(A+B)+\sin(A+B)\cos(A-B)}{1+\cos(A-B)} \\\\\\ = \frac{\sin(A+B)\cancel{(1+\cos(A-B))}}{\cancel{(1+\cos(A-B))}}\\\\\\ = \sin(A+B) \\\\\\ = \mathbb{LHS} \\\\\\ \mathcal{HENCE \quad PROVED}$

Answer 2

Answer:

HOPE THIS HELPS.......