Advertisements
Advertisements
प्रश्न
`5/(sin^2θ) - 5cot^2θ`, complete the activity given below.
Activity:
`5/(sin^2θ) - 5cot^2θ`
= `square (1/(sin^2θ) - cot^2θ)`
= `5(square - cot^2θ) ...[1/(sin^2θ) = square]`
= 5(1)
= `square`
Advertisements
उत्तर
`5/(sin^2θ) - 5cot^2θ`
= \[\boxed{5}\left(\frac{1}{\text{sin}^2θ} - \text{cot}^2θ\right)\]
= \[5\left(\boxed{\text{cosec}^2θ} - \text{cot}^2θ\right)\] \[...[\frac{1}{\text{sin}^2θ} = \boxed{\text{cosec}^2θ}]\]
= 5(1)
= \[\boxed{5}\]
APPEARS IN
संबंधित प्रश्न
Prove the following identities:
`secA/(secA + 1) + secA/(secA - 1) = 2cosec^2A`
Prove the following identities:
`(sinA + cosA)/(sinA - cosA) + (sinA - cosA)/(sinA + cosA) = 2/(2sin^2A - 1)`
Prove that:
(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1
`(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) = 2/((sin^2 theta - cos ^2 theta)) = 2/((2 sin^2 theta -1))`
If `(x/a sin a - y/b cos theta) = 1 and (x/a cos theta + y/b sin theta ) =1, " prove that "(x^2/a^2 + y^2/b^2 ) =2`
If \[\sin \theta = \frac{1}{3}\] then find the value of 9tan2 θ + 9.
sec4 A − sec2 A is equal to
cos4 A − sin4 A is equal to ______.
Prove the following identity :
`cosec^4A - cosec^2A = cot^4A + cot^2A`
Prove the following identity :
`sin^8θ - cos^8θ = (sin^2θ - cos^2θ)(1 - 2sin^2θcos^2θ)`
Find the value of x , if `cosx = cos60^circ cos30^circ - sin60^circ sin30^circ`
Prove that `sin(90^circ - A).cos(90^circ - A) = tanA/(1 + tan^2A)`
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
sec 60° = ?
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is ______.
sin(45° + θ) – cos(45° – θ) is equal to ______.
Show that tan4θ + tan2θ = sec4θ – sec2θ.
sec θ when expressed in term of cot θ, is equal to ______.
