Question

जर `1/sin^2θ - 1/cos^2θ-1/tan^2θ-1/cot^2θ-1/sec^2θ-1/("cosec"^2θ) = -3`, तर θ ची किमत काढा.

Answer 1

`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`

∴ `1/sin^2θ - 1/cos^2θ - 1/(((sin^2θ)/(cos^2θ))) - 1/(((cos^2θ)/(sin^2θ))) - cos^2θ - sin^2θ = -3`

∴ `1/sin^2θ - 1/cos^2θ - (cos^2θ)/(sin^2θ) - (sin^2θ)/(cos^2θ) - cos^2θ - sin^2θ = -3`

∴ `1/sin^2θ - (cos^2θ)/(sin^2θ) - 1/cos^2θ - (sin^2θ)/(cos^2θ) - (cos^2θ + sin^2θ) = -3`

∴ `(1 - cos^2θ)/(sin^2θ) - (1 + sin^2θ)/(cos^2θ) - 1 = -3`   ...(∵ sin²θ + cos²θ = 1)

∴ `sin^2θ/sin^2θ - (1 + sin^2θ)/cos^2θ - 1 = -3`   ...(∵ 1 − cos²θ = sin²θ)`

∴ `1 - (1 + sin^2θ)/cos^2θ - 1 = -3`

∴ `(1 + sin^2θ)/(1 - sin^2θ) = -3`

∴ `(1 + sin^2θ)/(1 - sin^2θ) = 3`   ...(दोन्ही बाजूंना −1 ने गुणून)

∴ 1 + sin²θ = 3 − 3sin²θ

∴ sin²θ + 3 sin²θ = 3 − 1

∴ 4sin²θ = 2

∴ `sin^2θ = 2/4`   ...(दोन्ही बाजूंना 4 ने गुणून)

∴ `sin^2θ = 1/2`

∴`sinθ = 1/sqrt2`   ...(दोन्ही बाजूंचे वर्गमूळ घेऊन)

∴ θ = 45°