Advertisements
Advertisements
प्रश्न
Find the value of x , if `cosx = cos60^circ cos30^circ - sin60^circ sin30^circ`
उत्तर
`cos(2x - 6)= cos^2 30^circ - cos^2 60^circ`
⇒ `cos(2x - 6) = cos^2(90^circ - 60^circ) - cos^2 60^circ`
⇒ `cos(2x - 6) = sin^2 60^circ - cos^2 60^circ`
⇒ `cos(2x - 6) = 1 - 2cos^2 60^circ = 1 - 2(1/2)^2 = 1 - 1/2 = 1/2`
⇒ `cos(2x - 6) = 1/2`
⇒ `cos(2x - 6) = cos60^circ`
⇒ `(2x - 6) = 60^circ`
⇒ `2x = 66^circ`
⇒ `x = 33^circ`
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`cos A/(1 - tan A) + sin A/(1 - cot A) = sin A + cos A`
If tan A = n tan B and sin A = m sin B, prove that:
`cos^2A = (m^2 - 1)/(n^2 - 1)`
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
`(sin theta+1-cos theta)/(cos theta-1+sin theta) = (1+ sin theta)/(cos theta)`
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Without using trigonometric table , evaluate :
`(sin49^circ/sin41^circ)^2 + (cos41^circ/sin49^circ)^2`
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)
If sec θ = `41/40`, then find values of sin θ, cot θ, cosec θ
Prove that cosec θ – cot θ = `sin theta/(1 + cos theta)`