JEE Main 20245 Apr, Morning shiftMathematicsEasyTrigonometry118× since 2002Suppose θ∈[0,π4]\theta \in\left[0, \frac{\pi}{4}\right]θ∈[0,4π] is a solution of 4cosθ−3sinθ=14 \cos \theta-3 \sin \theta=14cosθ−3sinθ=1. Then cosθ\cos \thetacosθ is equal to :AOption A. 6−6(36−2)\frac{6-\sqrt{6}}{(3 \sqrt{6}-2)}(36−2)6−6BOption B. 4(36+2)\frac{4}{(3 \sqrt{6}+2)}(36+2)4COption C. 6+6(36+2)\frac{6+\sqrt{6}}{(3 \sqrt{6}+2)}(36+2)6+6DOption D. 4(36−2)\frac{4}{(3 \sqrt{6}-2)}(36−2)4CheckSkip · show solution
