Question

The value of ${\sec }^2{10}^{\circ }+{cosec}^2{20}^{\circ }+{cosec}^2{40}^{\circ }$ is

Answer 1

Given : ${\sec }^2{10}^{\circ }+{cosec}^2{20}^{\circ }+{cosec}^2{40}^{\circ }$
$=\frac{1}{{\cos }^2{10}^{\circ }}+\frac{1}{{\sin }^2{20}^{\circ }}+\frac{1}{{\sin }^2{40}^{\circ }}=\frac{1}{{\cos }^2{10}^{\circ }}+\frac{4{\cos }^2{20}^{\circ }}{{\sin }^2{40}^{\circ }}+\frac{1}{{\sin }^2{40}^{\circ }}=\frac{1}{{\cos }^2{10}^{\circ }}+\frac{4{\cos }^2{20}^{\circ }+1}{{\sin }^2{40}^{\circ }}=\frac{1}{{\cos }^2{10}^{\circ }}+\frac{2(1+\cos {40}^{\circ })+1}{{\sin }^2{40}^{\circ }}=\frac{1}{{\cos }^2{10}^{\circ }}+\frac{(3+2\cos {40}^{\circ })4{\cos }^2{40}^{\circ }}{4{\cos }^2{40}^{\circ }{\sin }^2{40}^{\circ }}=\frac{1}{{\cos }^2{10}^{\circ }}+\frac{(3+2\cos {40}^{\circ })4{\cos }^2{40}^{\circ }}{{\sin }^2{80}^{\circ }}=\frac{1}{{\cos }^2{10}^{\circ }}+\frac{(3+2\cos {40}^{\circ })4{\cos }^2{40}^{\circ }}{{\sin }^2{80}^{\circ }}=\frac{1+(3+2\cos {40}^{\circ })2(1+\cos {80}^{\circ })}{{\cos }^2{10}^{\circ }}=\frac{1+2(3+2\cos {40}^{\circ }\cos {80}^{\circ }+3\cos {80}^{\circ }+2\cos {40}^{\circ })}{{\cos }^2{10}^{\circ }}=\frac{1+2(3+\cos {40}^{\circ }+\cos {120}^{\circ }+3\cos {80}^{\circ }+2\cos {40}^{\circ })}{{\cos }^2{10}^{\circ }}=\frac{6+6\cos {40}^{\circ }+6\cos {80}^{\circ }}{{\cos }^2{10}^{\circ }}=\frac{6+12\cos {60}^{\circ }\cos {20}^{\circ }}{{\cos }^2{10}^{\circ }}=\frac{6(1+\cos {20}^{\circ })}{{\cos }^2{10}^{\circ }}=12$