Inverse Trigonometric Functions

Given tan ^-1 $\frac{cosx-sinx}{cosx+sinx}$ , $\frac{-\pi }{4},$ x< $\frac{3x}{4}$

(M) Dividing numerator & denominator cos x we get,

tan ^-1 $\frac{\frac{cosx-sinx}{cosx}}{\frac{cosx+sinx}{cosx}}=ta{n}^{-1}\frac{\frac{cosx}{cosx}-\frac{sinx}{cosx}}{\frac{cosx}{cosx}+\frac{sinx}{cosx}}=ta{n}^{-1}\frac{1-tanx}{1+tanx.}$

We know that $tan\frac{\pi }{4}=1$ = 1 so,

$=ta{n}^{-1}\frac{tan\frac{\pi }{4}-tanx}{1+tan\frac{\pi }{4}tanx}=ta{n}^{-1}\left[tan\left(\frac{\pi }{4}-x\right)\right]$ $\{?tan(x-y)=\frac{tanx-tany}{1+tanx·tany}\}$

$=\frac{\pi }{4}-x$ .

L.H.S= 2 tan ^-1 $\frac{1}{2}$ + tan ^-1 $\frac{1}{7}$

(E) Using2 tan ^-1x= tan ^-1 $\frac{2a}{1-x^2}$ we can write.

L.H.S = tan ^-1 $\frac{2*\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{1-{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1 $\frac{1}{1\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$ + tan ^-1$\frac{1}{7}$

= tan ^-1 $\frac{1}{4\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$ + tan ^-1$\frac{1}{7}$ = tan ^-1 $\frac{4}{3}$ + tan ^-1 $\frac{1}{7}$

= tan ^-1$\frac{\frac{4}{3}+\frac{1}{7}}{1-\frac{4}{3}*\frac{1}{7}}$ { Ø tan -1 x + tan -1 y = tan - 1 $\frac{x+y}{1-xy}$ }

= tan ^-1 $\frac{\frac{7*4+1*3}{3*7}}{\frac{3*7-4*1}{3*7}}$

= tan -1 $\frac{28+3}{21-4}$ = tan ^-1 $\frac{31}{17}$ = R H S

L.H.S = tan ^-1$\frac{2}{11}$ + tan ^-1 $\frac{7}{24}$

Using tan ^-1x+ tan ^-1y= tan ^-1 $\frac{x+y}{1-xy}$ , xy<1

L.H.S =tan ^-1 $\frac{\frac{2}{11}+\frac{7}{24}}{1-\frac{2}{11}*\frac{7}{24}}$ = tan ^-1 $\frac{\frac{2*24+7*2}{11*24}}{\frac{11*24-7*2}{11*24}}$

tan ^-1 $\frac{48+14}{264-14}$ = tan ^-1 $\frac{125}{250}$ = tan ^-1 $\frac{1}{2}$ = R.H.S

We know that,

cos3θ= 4cos³θ - 3cosθ

Letx = cosθ Then θ = cos^-1x. We have,

Cos3 (cos^-1x) = 4x³-3x

3cos^-1x = cos^-1 (4x³- 3x)

Hence Proved

We know that.

sin 3θ =3 sin θ 4sin³θ (identity).

(E) Let x = sinθ. Then, sin ⁻¹x=θ . We have,

Sin3 (sin ⁻¹x) = 3x−4x³

3sin ⁻¹x =sin^-1 (3x−4x³)

Hence proved.

=tan⁻¹ $\left(\frac{tan\pi }{3}\right)$ − $\pi$ + sec⁻¹$\left(sec\frac{\pi }{3}\right)$

= $\frac{\pi }{3}-\pi +\frac{\pi }{3}$

= $\frac{\pi -3\pi +\pi }{3}$

= $\frac{-\pi }{3}.$

Option B is correct.

Given, Sin⁻¹x=y.

(E) We know that the principal value branch of Sin⁻¹ is

$[\frac{-\pi }{2},\frac{\pi }{2}]$ Hence,  $\frac{-\pi }{2}$ ≤ y ≤ $\frac{\pi }{2}$

Option B is correct.

cos⁻¹$\frac{1}{2}$ + 2Sin⁻¹ $\frac{1}{2}$ = cos⁻¹ $\left(cos\frac{\pi }{3}\right)$ + 2*Sin⁻¹$\left(sin\frac{\pi }{6}\right)$

= $\frac{\pi }{3}+2*\frac{\pi }{6}$

= $\frac{\pi }{3}+\frac{\pi }{3}$

= $\frac{2\pi }{3}$

tan⁻¹ (1) + cos⁻¹ $\left(\frac{1}{2}\right)$ +  sin ⁻¹$\left(\frac{-1}{2}\right)$

Let cos ^-1$\left(\frac{-1}{2}\right)$ =y Then cos y = $\frac{-1}{2}$ = − cos
$\frac{\mathrm{}\pi }{3}$$\frac{}{}$ cos $\left(\pi -\frac{x}{3}\right)$

= cos $\frac{3\pi -\pi }{3}$

= cos $\frac{2\pi }{3}$

$$  (E) We know that the range of principal value

branch of cos⁻¹ is [0, $\frac{\pi }{}$] and cos $\frac{2x}{3}$ = $\frac{-1}{2}$

Principal value of cos−1 $\left(\frac{-1}{2}\right)$ is $\frac{2x}{3}$