Suppose θ ∈ [0, π/4] is a solution of 4cosθ - 3sinθ = 1. Then cosθ is equal to

Question

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 :

(1) \(\frac{4}{(3 \sqrt{6}-2)}\)

(2) \(\frac{6-\sqrt{6}}{(3 \sqrt{6}-2)}\)

(3) \(\frac{6+\sqrt{6}}{(3 \sqrt{6}+2)}\)

(4) \(\frac{4}{(3 \sqrt{6}+2)}\)

Answer 1

Correct option is (1) \(\frac{4}{(3 \sqrt{6}-2)}\)

\(4\left(\frac{1-\tan ^{2} \theta / 2}{1+\tan ^{2} \theta / 2}\right)-3\left(\frac{2 \tan \frac{\theta}{2}}{1+\tan ^{2} \frac{\theta}{2}}\right)=1\)

let \(\tan \frac{\theta}{2}=\mathrm{t}\)

\(\frac{4-4 \mathrm{t}^{2}-6 \mathrm{t}}{1+\mathrm{t}^{2}}=1\)

\(4-4 \mathrm{t}^{2}-6 \mathrm{t}=1+\mathrm{t}^{2}\)

\(\Rightarrow 5 \mathrm{t}^{2}+6 \mathrm{t}-3=0\)

\(\Rightarrow \mathrm{t}=\frac{-6 \pm \sqrt{36-4(5)(-3)}}{2(5)}\)

\(=\frac{-6 \pm \sqrt{96}}{10} \)

\(=\frac{-6 \pm 4 \sqrt{6}}{10}\)

\(\mathrm{t}=\frac{-3+2 \sqrt{6}}{5} \)

\(\cos \theta=\frac{1-\mathrm{t}^{2}}{1+\mathrm{t}^{2}}=\frac{1-\left(\frac{2 \sqrt{6}-3}{5}\right)^{2}}{1+\left(\frac{2 \sqrt{6}-3}{5}\right)^{2}}=\frac{1-\left(\frac{24+9-12 \sqrt{6}}{25}\right)}{1+\left(\frac{24+9-12 \sqrt{6}}{25}\right)} \)

\( =\frac{25-33+12 \sqrt{6}}{25+33-12 \sqrt{6}}=\frac{12 \sqrt{6}-8}{58-12 \sqrt{6}}=\frac{6 \sqrt{6}-4}{29-6 \sqrt{6}} \times \frac{29+6 \sqrt{6}}{29+6 \sqrt{6}}\)

\(=\frac{100+150 \sqrt{6}}{625}=\frac{4+6 \sqrt{6}}{25} \times \frac{4-6 \sqrt{6}}{4-6 \sqrt{6}}\)

\(=\frac{-200}{25(4-6 \sqrt{6})}=\frac{-8}{4-6 \sqrt{6}}=\frac{4}{3 \sqrt{6}-2}\)