NIMCET 2027 Inverse Trigonometrical Function Previous Year Questions (PYQs) – Page 2 of 2

Let

$R = \sqrt{a^2 + b^2}$

Choose $\theta$ such that

$\cos \theta = \frac{b}{R}, \quad \sin \theta = \frac{a}{R}$

Then

$a\sin A + b\cos A = R(\sin A \sin \theta + \cos A \cos \theta) = R\cos(A - \theta)$

Given that

$a\sin A + b\cos A = c,$

we get

$R\cos(A - \theta) = c \quad \Rightarrow \quad \cos(A - \theta) = \frac{c}{\sqrt{a^2 + b^2}}$

Hence,

$A - \theta = \pm \cos^{-1}\!\left(\frac{c}{\sqrt{a^2 + b^2}}\right) + 2\pi n, \quad n \in \mathbb{Z}$

and since $\tan \theta = \dfrac{a}{b}$, we have $\theta = \tan^{-1}\!\left(\dfrac{a}{b}\right)$

Therefore,

$\boxed{A = \tan^{-1}\!\left(\frac{a}{b}\right) \pm \cos^{-1}\!\left(\frac{c}{\sqrt{a^2 + b^2}}\right)}$

Given,

$PA=60$

$PB=80$

$\angle APB=60^\circ$

Using cosine rule in triangle $APB$,

$AB^2=PA^2+PB^2-2(PA)(PB)\cos 60^\circ$

$AB^2=60^2+80^2-2(60)(80)\cdot \frac{1}{2}$

$AB^2=3600+6400-4800$

$AB^2=5200$

$AB=\sqrt{5200}$

$AB=\sqrt{400\cdot 13}$

$AB=20\sqrt{13}$

Given,

$\tan^{-1}(3x)+\tan^{-1}(2x)=\frac{\pi}{4}$

Taking tangent on both sides,

$\tan\left(\tan^{-1}(3x)+\tan^{-1}(2x)\right)=\tan\frac{\pi}{4}$

Using,

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

we get,

$\frac{3x+2x}{1-(3x)(2x)}=1$

$\frac{5x}{1-6x^2}=1$

So,

$5x=1-6x^2$

$6x^2+5x-1=0$

Factorizing,

$6x^2+6x-x-1=0$

$6x(x+1)-1(x+1)=0$

$(x+1)(6x-1)=0$

So,

$x=-1$ or $x=\frac{1}{6}$

Now check both values.

For $x=\frac{1}{6}$,

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

So, $x=\frac{1}{6}$ is valid.

For $x=-1$,

$\tan^{-1}(-3)+\tan^{-1}(-2)$ is negative, so it cannot be equal to $\frac{\pi}{4}$.

Therefore, only one solution exists.