If tan θ₁ tan θ₂ = k, Find cos(θ₁ − θ₂) / cos(θ₁ + θ₂)

If tan θ₁ tan θ₂ = k, Find cos(θ₁ − θ₂) / cos(θ₁ + θ₂)

Question:
If \[ \tan\theta_1\tan\theta_2=k \] then \[ \frac{\cos(\theta_1-\theta_2)} {\cos(\theta_1+\theta_2)} \] is equal to
(a) \(\frac{1+k}{1-k}\)
(b) \(\frac{1-k}{1+k}\)
(c) \(\frac{k+1}{k-1}\)
(d) \(\frac{k-1}{k+1}\)

Solution

Using the identities:

\[ \cos(A-B)=\cos A\cos B+\sin A\sin B \]

and

\[ \cos(A+B)=\cos A\cos B-\sin A\sin B \]

Therefore,

\[ \frac{\cos(\theta_1-\theta_2)} {\cos(\theta_1+\theta_2)} = \frac{ \cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2 } { \cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2 } \]

Dividing numerator and denominator by

\[ \cos\theta_1\cos\theta_2 \]

we get

\[ = \frac{ 1+\tan\theta_1\tan\theta_2 } { 1-\tan\theta_1\tan\theta_2 } \]

Given,

\[ \tan\theta_1\tan\theta_2=k \]

Substituting,

\[ = \frac{1+k}{1-k} \]

Final Answer

\[ \boxed{ \frac{\cos(\theta_1-\theta_2)} {\cos(\theta_1+\theta_2)} = \frac{1+k}{1-k} } \]

Correct Option: (a)

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