**(a)** If $ an heta + \dfrac{1}{ an heta} = 2$, find the value...

Question

**(a)** If $	an	heta + \dfrac{1}{	an	heta} = 2$, find the value of $	an^2	heta + \dfrac{1}{	an^2	heta}$.

**OR**

**(b)** Prove that : $\sqrt{\dfrac{1-\sin	heta}{1+\sin	heta}} = \sec	heta - 	an	heta$

Accepted Answer

### (a)

**Step 1 — Square both sides of the given equation.**
$$\left(	an	heta + \frac{1}{	an	heta}ight)^2 = 2^2$$

**Step 2 — Expand using $(a+b)^2 = a^2 + b^2 + 2ab$.**
$$	an^2	heta + \frac{1}{	an^2	heta} + 2\left(	an	heta \cdot \frac{1}{	an	heta}ight) = 4$$

**Step 3 — Simplify the middle term.**
$$	an^2	heta + \frac{1}{	an^2	heta} + 2 = 4$$

**Step 4 — Solve.**
$$	an^2	heta + \frac{1}{	an^2	heta} = 4 - 2 = 2$$

$$\boxed{	an^2	heta + \frac{1}{	an^2	heta} = 2}$$

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### (b)

**Step 1 — Take the LHS and rationalise by multiplying numerator and denominator by $(1 - \sin	heta)$.**
$$	ext{LHS} = \sqrt{\frac{1-\sin	heta}{1+\sin	heta} 	imes \frac{1-\sin	heta}{1-\sin	heta}}$$

**Step 2 — Simplify.**
$$= \sqrt{\frac{(1-\sin	heta)^2}{1-\sin^2	heta}}$$

**Step 3 — Use the identity $1 - \sin^2	heta = \cos^2	heta$.**
$$= \sqrt{\frac{(1-\sin	heta)^2}{\cos^2	heta}} = \frac{1-\sin	heta}{\cos	heta}$$

**Step 4 — Split the fraction.**
$$= \frac{1}{\cos	heta} - \frac{\sin	heta}{\cos	heta} = \sec	heta - 	an	heta = 	ext{RHS}$$

$$	extbf{Hence proved.}$$

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