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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.
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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संबंधित प्रश्न
Prove the following trigonometric identities.
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`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
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