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प्रश्न
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
उत्तर
LHS = `sqrt(2 + tan^2 θ + cot^2 θ)`
= `sqrt( tan^2 θ + cot^2θ + 2tan θ.cot θ)` ...[ ∵ tan θ.cot θ = 1 ]
= `sqrt( tan^2 θ + cot^2θ)`
= tan θ + cot θ
= RHS
Hence proved.
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संबंधित प्रश्न
Prove the following trigonometric identities.
sec6θ = tan6θ + 3 tan2θ sec2θ + 1
Prove the following identities:
`cot^2A((secA - 1)/(1 + sinA)) + sec^2A((sinA - 1)/(1 + secA)) = 0`
`sin^6 theta + cos^6 theta =1 -3 sin^2 theta cos^2 theta`
Prove the following identity :
`cosec^4A - cosec^2A = cot^4A + cot^2A`
Prove the following identity :
`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
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`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
For ΔABC , prove that :
`tan ((B + C)/2) = cot "A/2`
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
`5/(sin^2theta) - 5cot^2theta`, complete the activity given below.
Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
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`
