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प्रश्न
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
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उत्तर
LHS = `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ`
= `(sin^4θ + cos^4θ)/(sin^2θ.cos^2θ)`
= `((sin^2 θ + cos^2 θ)^2 - 2(sin^2 θ. cos^2 θ))/(sin^2 θ.cos^2 θ)`
= `((1)^2 - 2sin^2θ. cos^2 θ)/(sin^2 θ.cos^2 θ)`
= `1/(sin^2 θ.cos^2 θ) - (2sin^2θ. cos^2 θ)/(sin^2 θ.cos^2 θ)`
= `1/(sin^2 θ.cos^2 θ) - 2`
= RHS
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`tan θ/(1 - cot θ) + cot θ/(1 - tan θ) = 1 + tan θ + cot θ`
Prove that:
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
` (sin theta - cos theta) / ( sin theta + cos theta ) + ( sin theta + cos theta ) / ( sin theta - cos theta ) = 2/ ((2 sin^2 theta -1))`
Prove the following identities:
`(cos theta "cosec" theta - sin theta sec theta )/(cos theta + sin theta) = "cosec" theta - sec theta`
If \[\sin \theta = \frac{1}{3}\] then find the value of 2cot2 θ + 2.
Prove that `((1 - cos^2 θ)/cos θ)((1 - sin^2θ)/(sin θ)) = 1/(tan θ + cot θ)`
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
sec θ when expressed in term of cot θ, is equal to ______.
