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 trigonometric identities.
`(cot A - cos A)/(cot A + cos A) = (cosec A - 1)/(cosec A + 1)`
Prove the following identities:
sec2A + cosec2A = sec2A . cosec2A
Prove the following identities:
`cosecA + cotA = 1/(cosecA - cotA)`
Prove the following identities:
sec2 A . cosec2 A = tan2 A + cot2 A + 2
Prove the following identities:
cosec4 A (1 – cos4 A) – 2 cot2 A = 1
If 2 sin A – 1 = 0, show that: sin 3A = 3 sin A – 4 sin3 A
If 4 cos2 A – 3 = 0, show that: cos 3 A = 4 cos3 A – 3 cos A
`cos^2 theta + 1/((1+ cot^2 theta )) =1`
Show that none of the following is an identity:
`cos^2theta + cos theta = 1`
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that (x2 – y2) = (a2 – b2).
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
If sin θ + sin2 θ = 1, then cos2 θ + cos4 θ =
The value of sin ( \[{45}^° + \theta) - \cos ( {45}^°- \theta)\] is equal to
Prove the following identity :
tanA+cotA=secAcosecA
Prove the following identity :
`(secA - 1)/(secA + 1) = (1 - cosA)/(1 + cosA)`
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
If sec θ = `25/7`, then find the value of tan θ.
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2
If 4 tanβ = 3, then `(4sinbeta-3cosbeta)/(4sinbeta+3cosbeta)=` ______.
