Question

sin(cot^-1 3/4)=cos(tan^-1 x) find x

Answer 1

cos(tan−1x)=sin[cot−1(34)]      ...(1)As we know that sinx=cos(π2−x),sosin[cot−1(34)] =cos[π2−cot−1(34)]Now we know that tan−1x+cot−1x=π2 soπ2−cot−1(34)=tan−1(34) putting this, we getsin[cot−1(34)]=cos(π2−cot−1(34))=cos[tan−1(34)]Now, putting this value in eq(1), we getcos(tan−1x)=cos[tan−1(34)]⇒tan−1x=tan−1(34)⇒x=34

Answer 2

Answer:

$x=\frac{4}{3}$

Step-by-step explanation:

$cos(tan^{-1}x)=sin(cot^{-1}\frac{3}{4})$

Let$cot^{-1}\frac{3}{4} = y$

$cot y =\frac{3}{4}$

$Cot y = \frac{perpendicular}{base}$

On comaring

Perpendicular = 3

Base = 4

Using Pythagoras's theorem,

$hypotenuse=\sqrt{p^2+b^2}=\sqrt{3^2+4^2}=\sqrt{25}=5$

$siny=\frac{P}{B}=\frac{3}{5}$

$sin(cot^{-1}\frac{3}{4}=siny=\frac{3}{5}$

$cos(tan^{-1}x)=3/5$

$tan^{-1}x=cos^{-1}(3/5)$

$cos^{-1}(3/5)=z$

$cosz=3/5$

perpendicular=√5²-3²=√16=4

tanz==4/3

tan⁻¹x=z

x=tanz

$x=\frac{4}{3}$