If $$\sin\theta=a\cos\phi $$ and $$\cos\theta=b\sin\phi$$, then the value of $$(a^{2}-1)\cot^{2}\phi+(1-b^{2})\cot^{2}\theta$$ is equal to:

Given : $$\sin\theta=a\cos\phi $$ and $$\cos\theta=b\sin\phi$$ ---------(i)

To find : $$(a^{2}-1)cot^{2}\phi+(1-b^{2})cot^{2}\theta$$

= $$(a^{2}-1)\frac{cos^{2}\phi}{sin^2\phi}+(1-b^{2})\frac{cos^{2}\theta}{sin^2\theta}$$

= $$\frac{(a^2-1)cos^2\phi.sin^2\theta+(1-b^2)cos^2\theta.sin^2\phi}{sin^2\phi.sin^2\theta}$$

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

Substituting value from equation (i), we get :

= $$\frac{sin^2\theta.sin^2\theta-cos^2\phi.sin^2\theta+cos^2\theta.sin^2\phi-cos^2\theta.cos^2\theta}{sin^2\phi.sin^2\theta}$$

= $$\frac{sin^4\theta-cos^4\theta-cos^2\phi.sin^2\theta+cos^2\theta.sin^2\phi}{sin^2\phi.sin^2\theta}$$

= $$\frac{(sin^2\theta-cos^2\theta)(sin^2\theta+cos^2\theta)-cos^2\phi.sin^2\theta+cos^2\theta.sin^2\phi}{sin^2\phi.sin^2\theta}$$

Using, $$(sin^2\theta+cos^2\theta=1)$$

= $$\frac{sin^2\theta-cos^2\theta-cos^2\phi.sin^2\theta+cos^2\theta.sin^2\phi}{sin^2\phi.sin^2\theta}$$

= $$\frac{sin^2\theta-cos^2\phi.sin^2\theta-cos^2\theta+cos^2\theta.sin^2\phi}{sin^2\phi.sin^2\theta}$$

= $$\frac{sin^2\theta(1-cos^2\phi)-cos^2\theta(1-sin^2\phi)}{sin^2\phi.sin^2\theta}$$

= $$\frac{sin^2\theta(sin^2\phi)-cos^2\theta(cos^2\phi)}{sin^2\phi.sin^2\theta}$$

= $$1-(\frac{cos^2\theta}{sin^2\phi})(\frac{cos^2\phi}{sin^2\theta})$$

= $$1-\frac{b^2}{a^2}$$

= $$\frac{a^{2}-b^{2}}{a^{2}}$$

=> Ans - (D)