मराठी
महाराष्ट्र राज्य शिक्षण मंडळएस.एस.सी (इंग्रजी माध्यम) इयत्ता १० वी

Θθθθθcosecθ1sin2θ-1cos2θ-1tan2θ-1cot2θ-1sec2θ-1cosec2θ=-3, then find the value of θ.

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प्रश्न

`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.

बेरीज
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उत्तर

`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°

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