The general solution of cot θ+tan θ=2 is

Correct answer: (B) (n I)/2+(-1)^n n/4. Given that, cot θ+tan θ=2 ⇒ (cos θ)/(sin θ)+(sin θ)/(cos θ)=2 ⇒ fraccos^2 θ+sin^2 θsin θ cos θ=2 ⇒ 1/(sin θ cos θ)=2 ⇒ 2 sin θ cos θ=1 ⇒ sin 2 θ=1=sin ((Π)/2 ) As we know, if sin θ=sin α then θ=n Π+(-1)^n α Therefore, 2 θ=n Π+(-1)^n (Π)/2 ⇒ θ=(n Π)/2+(-1)^n (Π)/4

Q79 / 180 · KCET 2016 · Mathematics

The general solution of $ \cot \theta+\tan \theta=2 $ is

A
$ \theta=\frac{n \Pi}{2}+(-1)^{n \frac{n I}{8}} $
Correct answer
B
$ \frac{n I}{2}+(-1)^{n \frac{n}{4}} $
C
$ \theta=\frac{n \Pi}{2}+(-1)^{n} \frac{n}{6} $
D
$ \theta=n \Pi+(-1)^{n} \frac{n}{8} $

Solution

Given that, $\cot \theta+\tan \theta=2$ $\Rightarrow \frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}=2$ $\Rightarrow \frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}=2$ $\Rightarrow \frac{1}{\sin \theta \cos \theta}=2$ $\Rightarrow 2 \sin \theta \cos \theta=1$ $\Rightarrow \sin 2 \theta=1=\sin \left(\frac{\Pi}{2}\right)$ As we know, if $\sin \theta=\sin \alpha$ then $\theta=n \Pi+(-1)^{n} \alpha$ Therefore, $2 \theta=n \Pi+(-1)^{n} \frac{\Pi}{2}$ $\Rightarrow \theta=\frac{n \Pi}{2}+(-1)^{n} \frac{\Pi}{4}$

Hence, the correct answer is option: 'B'.

The general solution of \( \cot \theta+\tan \theta=2 \) is

Solution

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