JEE Mains · Maths · STD 11 - 3. trignometrical ratios,functions and identities
Suppose \(\theta \in\left[0, \frac{\pi}{4}\right]\) is a solution of \(4 \cos \theta-3 \sin \theta=1\) Then \(\cos \theta\) is equal to :
- A \(\frac{4}{(3 \sqrt{6}-2)}\)
- B \(\frac{6-\sqrt{6}}{(3 \sqrt{6}-2)}\)
- C \(\frac{6+\sqrt{6}}{(3 \sqrt{6}+2)}\)
- D \(\frac{4}{(3 \sqrt{6}+2)}\)
Answer & Solution
Correct Answer
(A) \(\frac{4}{(3 \sqrt{6}-2)}\)
Step-by-step Solution
Detailed explanation
\(4\left(\frac{1-\tan ^2 \theta / 2}{1+\tan ^2 \theta / 2}\right)-3\left(\frac{2 \tan \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}\right)=1\) let \(\tan \frac{\theta}{2}=\mathrm{t}\) \( \frac{4-4 t^2-6 t}{1+t^2}=1 \) \( 4-4 t^2-6 t=1+t^2\) \( \Rightarrow 5 t^2+6 t-3=0 \)…
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