Question 143

$$tan 9^{\circ}=\frac{p}{q}$$ then the value of $$\frac{\sec^281^{\circ}}{1+\cot^281^{\circ}}$$ is

Solution

we need to find value of $$\frac{\sec^281^{\circ}}{1+\cot^281^{\circ}}$$

given that : $$tan 9$$ = $$\frac{p}{q}$$

Using : 1 + $$tan^2 \theta$$ = $$sec^2 \theta$$ and $$cot^2 \theta$$ = $$tan^2 \theta$$

$$\frac{\sec^281^{\circ}}{1+\cot^281^{\circ}}$$

= $$\frac{1+tan^2 81}{1+tan^2 81}$$ x $$tan^2 81$$

= $$tan^2 81$$ = $$tan^2 (90-9)$$ = $$cot^2 9$$ = $$\frac{q^2}{p^2}$$

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