Advertisements
Advertisements
प्रश्न
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
Advertisements
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
`sec theta (1- sin theta )( sec theta + tan theta )=1`
` (sin theta + cos theta )/(sin theta - cos theta ) + ( sin theta - cos theta )/( sin theta + cos theta) = 2/ ((1- 2 cos^2 theta))`
If a cos `theta + b sin theta = m and a sin theta - b cos theta = n , "prove that "( m^2 + n^2 ) = ( a^2 + b^2 )`
Prove that secθ + tanθ =`(costheta)/(1-sintheta)`.
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
Prove the following identity :
`sqrt(cosec^2q - 1) = "cosq cosecq"`
Find the value of ( sin2 33° + sin2 57°).
Prove that : `(sin(90° - θ) tan(90° - θ) sec (90° - θ))/(cosec θ. cos θ. cot θ) = 1`
Prove that cot2θ × sec2θ = cot2θ + 1
(1 – cos2 A) is equal to ______.
