Question

`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.

Answer 1

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

cosec²θ − sec²θ  − cot²θ  − tan²θ − cos²θ − sin²θ  = −3   ...`[because sintheta = 1/(cosectheta), costheta = 1/(sectheta), tantheta = 1/(cottheta)]`

⇒ 1 + cot²θ − 1 − tan²θ − cot²θ − tan²θ − 1 = −3

⇒ − 2 tan²θ − 1 = − 3   ...`[(because 1 + cot^2theta = cosec^2theta), (1 + sec^2theta = tan^2theta), (sin^2theta + cos^2theta = 1)]`

⇒ −2 tan²θ = − 3 + 1

⇒ −2 tan²θ = −2

⇒ tan²θ = 1

⇒ tan θ = 1   ...(Taking square root on both sides)

⇒ tan θ = tan 45°

∴ θ = 45°