Question

Triangle PQR is an isosceles right triangle, right angled at Q. Find the value of sec² P + cosec² R.

Answer 1

Given: Triangle PQR is an isosceles right triangle, right angled at Q, so PQ = QR.

In a right isosceles triangle with right angle at Q, the two acute angles are equal and sum to 90°, so ∠P = ∠R = 45°.

sec² P + cosec² R = sec²(45°) + cosec²(45°).

Using `sec = 1/cos` and `"cosec" = 1/sin`.

cos 45° = sin 45° = `sqrt(2)/2`, so `sec 45^circ = sqrt(2)` and `"cosec"  45^circ = sqrt(2)`.

`sec^2(45^circ) = (sqrt(2))^2 = 2` and `"cosec"^2(45^circ) = (sqrt(2))^2 = 2`.

Therefore, sec² P + cosec² R = 2 + 2 = 4.