NIMCET 2008 — Mathematics PYQ

Since $\theta$ is the arithmetic mean of $\alpha$ and $\beta$, we have:

2. Apply Sum-to-Product Formulas

Given the first equation:

Using the identity $\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$:

Given the second equation:

Using the identity $\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$:

3. Find $\tan \theta$

Divide Equation 2 by Equation 1:

4. Express $\sin 2\theta$ and $\cos 2\theta$ in terms of $\tan \theta$

Using standard trigonometric double-angle identities:

• $\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }=\frac{2(b/a)}{1+(b/a{)}^2}=\frac{2b/a}{(a^2+b^2)/a^2}=\frac{2ab}{a^2+b^2}$

• $\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }=\frac{1-(b/a{)}^2}{1+(b/a{)}^2}=\frac{(a^2-b^2)/a^2}{(a^2+b^2)/a^2}=\frac{a^2-b^2}{a^2+b^2}$

5. Calculate $\sin 2\theta +\cos 2\theta$

Looking at the options:

• Option (a): $\frac{(a+b{)}^2}{a^2+b^2}=\frac{a^2+b^2+2ab}{a^2+b^2}$ (Incorrect)

• Option (b): $\frac{(a-b{)}^2}{a^2+b^2}=\frac{a^2+b^2-2ab}{a^2+b^2}$ (Incorrect)

• Option (c): $\frac{a^2-b^2}{a^2+b^2}$ (Incorrect)

Since the resulting expression $\frac{a^2-b^2+2ab}{a^2+b^2}$ does not exactly match any of the provided simplified forms in (a), (b), or (c):

Correct Option: (d) None of these