The problem is thus treated in steps:
We are given that:
cos(π/4 + θ) + cos(π/4 - θ) = √2 cosθ
We shall consider the left-hand side (LHS):
The sum and difference formulas for cosine are useful here:
cos(A + B) = cosA * cosB - sinA * sinB
cos(A - B) = cosA * cosB + sinA * sinB
Substituting into our equation we get:
LHS = [cos(π/4) * cosθ - sin(π/4) * sinθ] + [cos(π/4) * cosθ + sin(π/4) * sinθ]
Now we simplify:
LHS = 2 cos(π/4) cosθ
cos(π/4) = 1/√2. Thus:
LHS = 2(1/√2) cosθ
LHS = √2 cosθ.
Hence, LHS = RHS.
So we have proved that the given equation stands correct.
Question : $\frac{\cos \theta}{\sec \theta-1}+\frac{\cos \theta}{\sec \theta+1}$ is equal to :
Option 1: $2 \sec ^2 \theta$
Option 2: $2 \cot ^2 \theta$
Option 3: $2 \cos ^2 \theta$
Option 4: $2 \sin ^2 \theta$
Question : The value of $\frac{\sin\theta-2\sin^{3}\theta}{2\cos^{3}\theta-\cos\theta}$ is equal to:
Option 1: $\sin\theta$
Option 2: $\cos\theta$
Option 3: $\tan\theta$
Option 4: $\cot\theta$
Question : What is the value of $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}+\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}$?
Option 1: $\frac{1}{\left(\sin ^2 \theta-\cos ^2 \theta\right)}$
Option 2: $2\left(\sin ^2 \theta-\cos ^2 \theta\right)$
Option 3: $\frac{2}{\left(\sin ^2 \theta-\cos ^2 \theta\right)}$
Option 4: $\sin ^2 \theta-\cos ^2 \theta$
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