Advertisements
Advertisements
प्रश्न
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
Advertisements
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,
\[\cot^2 \theta - \frac{1}{\sin^2 \theta} = - 1\]
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`(i) cos4^4 A – cos^2 A = sin^4 A – sin^2 A`
`(ii) cot^4 A – 1 = cosec^4 A – 2cosec^2 A`
`(iii) sin^6 A + cos^6 A = 1 – 3sin^2 A cos^2 A.`
Prove the following trigonometric identities.
`(1 + tan^2 A) + (1 + 1/tan^2 A) = 1/(sin^2 A - sin^4 A)`
Prove the following identities:
`(1 + sin A)/(1 - sin A) = (cosec A + 1)/(cosec A - 1)`
Prove the following identities:
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove that:
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Prove the following identities:
`(1 - cosA)/sinA + sinA/(1 - cosA)= 2cosecA`
Prove that:
`cosA/(1 + sinA) = secA - tanA`
`1+((tan^2 theta) cot theta)/(cosec^2 theta) = tan theta`
If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`.
If `sqrt(3) sin theta = cos theta and theta ` is an acute angle, find the value of θ .
\[\frac{1 + \tan^2 A}{1 + \cot^2 A}\]is equal to
If cos \[9\theta\] = sin \[\theta\] and \[9\theta\] < 900 , then the value of tan \[6 \theta\] is
Prove the following identity :
`sqrt((1 + sinq)/(1 - sinq)) + sqrt((1- sinq)/(1 + sinq))` = 2secq
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
Prove that:
tan (55° + x) = cot (35° – x)
Prove that `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ = 3`.
Prove that `tan A/(1 + tan^2 A)^2 + cot A/(1 + cot^2 A)^2 = sin A.cos A`
Prove that : `tan"A"/(1 - cot"A") + cot"A"/(1 - tan"A") = sec"A".cosec"A" + 1`.
If cos (α + β) = 0, then sin (α – β) can be reduced to ______.
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
