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प्रश्न
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
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उत्तर
We have,
`cot ^2 -1/ sin^2 θ= cot ^2 θ-(1/ sinθ)^2`
= `cot ^2 θ-(cosec θ)^2`
= `cot^2 θ-cosec^2 θ`
We know that, `cot^2 θ-cosec^2 θ`
Therefore,
\[\cot^2 \theta - \frac{1}{\sin^2 \theta} = - 1\]
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संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
`(sintheta - 2sin^3theta)/(2costheta - costheta) =tan theta`
Show that `sqrt((1-cos A)/(1 + cos A)) = sinA/(1 + cosA)`
Prove the following trigonometric identity:
`sqrt((1 + sin A)/(1 - sin A)) = sec A + tan A`
Prove the following trigonometric identities.
`1 + cot^2 theta/(1 + cosec theta) = cosec theta`
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
`sec theta (1- sin theta )( sec theta + tan theta )=1`
`sqrt((1+cos theta)/(1-cos theta)) + sqrt((1-cos theta )/(1+ cos theta )) = 2 cosec theta`
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
The value of sin2 29° + sin2 61° is
Prove the following identity :
`sec^2A.cosec^2A = tan^2A + cot^2A + 2`
Without using trigonometric identity , show that :
`tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ = 1/sqrt(3)`
Find A if tan 2A = cot (A-24°).
Prove that `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ = 3`.
Prove that:
`(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(2 sin^2 A - 1)`
Without using the trigonometric table, prove that
tan 10° tan 15° tan 75° tan 80° = 1
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove that `(1 + sin θ)/(1 - sin θ) = (sec θ + tan θ)^2`.
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`
