If tan θ=ab, prove that a sin θ+b cos θa sin θ+b cos θ=a2−b2a2+b2.

Solution:

Given:

$\tan \theta=\frac{a}{b}$

Now, we know that $\tan \theta=\frac{\sin \theta}{\cos \theta}$

Therefore equation (1) becomes

$\frac{\sin \theta}{\cos \theta}=\frac{a}{b}$....(2)

Now, multiplying by $\frac{a}{b}$ on both sides of equation (2)

We get,

$\frac{a}{b} \times \frac{\sin \theta}{\cos \theta}=\frac{a}{b} \times \frac{a}{b}$

Therefore,

$\frac{a \sin \theta}{b \cos \theta}=\frac{a^{2}}{b^{2}}$...(3)

Now by applying dividendo in above equation (3)

We get,

$\frac{a \sin \theta-b \cos \theta}{b \cos \theta}=\frac{a^{2}-b^{2}}{b^{2}}$

Now by applying componendo in equation (3)

We get,

$\frac{a \sin \theta+b \cos \theta}{b \cos \theta}=\frac{a^{2}+b^{2}}{b^{2}}$....(5)

Now, by dividing equation (4) by equation (5)

We get,

$\frac{\frac{a \sin \theta-b \cos \theta}{b \cos \theta}}{\frac{a \sin \theta+b \cos \theta}{b \cos \theta}}=\frac{\frac{a^{2}-b^{2}}{b^{2}}}{\frac{a^{2}+b^{2}}{b^{2}}}$

Therefore,

$\frac{a \sin \theta-b \cos \theta}{b \cos \theta} \times \frac{b \cos \theta}{a \sin \theta+b \cos \theta}=\frac{a^{2}-b^{2}}{b^{2}} \times \frac{b^{2}}{a^{2}+b^{2}}$

Therefore, $b \cos \theta$ and $b^{2}$ cancels on L.H.S and R.H.S respectively and we get,

$\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$

Hence, it is proved that

$\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}=\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$