125
Triangle PQR is an isosceles right triangle, right angled at Q. Find the value of $\sec^2 P + \text{cosec}^2 R$.
Show SolutionHide Solution↓
Solution: (a) $\Delta PQR$ is an isosceles right triangle
$:: \angle P= \angle R = 45^{\circ}$ (1/2 Mark)
$\sec^2 P + \text{cosec}^2 R = \sec^2 45^{\circ} + \text{cosec}^2 45^{\circ}$
$= (\sqrt{2})^2 + (\sqrt{2})^2$
$= 4$ (1/2 Mark)
$:: \angle P= \angle R = 45^{\circ}$ (1/2 Mark)
$\sec^2 P + \text{cosec}^2 R = \sec^2 45^{\circ} + \text{cosec}^2 45^{\circ}$
$= (\sqrt{2})^2 + (\sqrt{2})^2$
$= 4$ (1/2 Mark)