Question

If $\tan \left(\pi cos\theta \right)=\cot \left(\pi \sin \theta \right),$ then $\cos (\theta -(\pi /4\left)\right)=\pm \left(1/\right(2\sqrt{2}\left)\right)$.

• A. True
• B. False

Answer 1

The correct option is A True

$tan\left(\pi cos\theta \right)=cot\left(\pi sin\theta \right)$

$tan\left(\pi cos\theta \right)=\frac{1}{tan\left(\pi sin\theta \right)}$

$tan\left(\pi cos\theta \right)\cdot tan\left(\pi sin\theta \right)=1$ ....................... (1)

Now,

We know that,

$tan(A+B)=\frac{tan\left(A\right)+tan\left(B\right)}{1-tan\left(A\right)\cdot tan\left(B\right)}$

$tan(\pi cos\theta +\pi sin\theta )=\frac{tan\left(\pi cos\theta \right)+tan\left(\pi sin\theta \right)}{1-tan\left(\pi cos\theta \right)\cdot tan\left(\pi sin\theta \right)}$

From (1), $tan\left(\pi cos\theta \right)\cdot tan\left(\pi sin\theta \right)=1$

$tan(\pi cos\theta +\pi sin\theta )=\frac{tan\left(\pi cos\theta \right)+tan\left(\pi sin\theta \right)}{1-1}$

$tan(\pi cos\theta +\pi sin\theta )=\pm \infty$

$\Rightarrow \pi cos\theta +\pi sin\theta =\pm \frac{\pi }{2}$

$cos\theta +sin\theta =\pm \frac{1}{2}$

Divide both sides by $\sqrt{2},$

$\frac{1}{\sqrt{2}}cos\theta +\frac{1}{\sqrt{2}}sin\theta =\pm \frac{1}{2\sqrt{2}}$

$cos\frac{\pi }{4}.cos\theta +sin\frac{\pi }{4}sin\theta =\pm \frac{1}{2\sqrt{2}}$

$cos(\theta -\frac{\pi }{4})=\pm \frac{1}{2\sqrt{2}}$

So, the given statement is true.