Hi Sanjith, Here is the answer to your question. Given: 2 sin 6 θ cos 6 θ − 3 sin 4 θ cos 4 θ + 1 = 0 Let x = sin 2 θ cos 2 θ Then: 2 x 3 − 3 x 2 + 1 = 0 Factorising: ( x − 1 ) ( 2 x 2 − x − 1 ) = 0 So, x = 1 or x = 1 or x = − 1 2 . Since x = sin 2 θ cos 2 θ ≥ 0 , valid value is x = 1 .
Question : If $\sin A+\sin ^2 A=1$, then the value of $\cos ^4 A+\cos ^6 A$ is:
Option 1: $\cos A$
Option 2: $\sin A$
Option 3: 1
Option 4: 0
Question : If $6 \cot \theta=5$, then find the value of $\frac{(6 \cos \theta+\sin \theta)}{(6 \cos\theta-4 \sin\theta)}$
Option 1: 5
Option 2: 1
Option 3: 6
Question : If $\cos A + \cos B + \cos C = 3$, then what is the value of $\sin A + \sin B + \sin C$?
Option 1: $1$
Option 2: $2$
Option 3: $0$
Option 4: $-1$
Question : If $\cos x+\sin x=\sqrt{2} \cos x$, what is the value of $(\cos x-\sin x)^2+(\cos x+\sin x)^2$?
Option 1: $2$
Option 2: $1$
Option 4: $\frac{1}{\sqrt{2}}$
Question : $\frac{\sin^4 \theta+\cos^4 \theta}{1-2 \sin^2 \theta \cos^2 \theta}=$____.
Option 1: 1
Option 2: 2
Option 3: – 1
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