Question

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

Answer 1

Answer:

x = ± 3/√7

Step-by-step explanation:

$\rm cos(tan^{ - 1} x) = sin \bigg(cos^{ - 1} \dfrac{3}{4} \bigg)$

Let tan^( - 1 ) x = a

tan a = x = x/1 = P/B

• P = x
• B = 1
• H = ?

By Pythagorean theorem

H² = P² + B²

H² = x² + 1²

H = √( x² + 1 )

$\rm Now \ \ LHS = cos(tan^{ - 1} x)$

= cos a = B/H = 1/√( x² + 1 )

Let cos^( - 1 ) 3/4 = b

cos b = 3/4 = B/H

• P = ?
• B = 3
• H = 4

By Pythagorean theorem

H² = P² + B²

4² = P² + 3²

16 = P² + 9

P² = 16 - 9

P² = 7

P = √7

$\rm Now \ \ RHS = sin \bigg(cos^{ - 1} \dfrac{3}{4} \bigg)$

= sin b = P/H = √7 /4

Now equate LHS and RHS of the equation

$\rm \dfrac{1}{ \sqrt{x^2 + 1 } } = \dfrac{ \sqrt{7} }{4}$

$\rm \Rightarrow 4 = \sqrt{7( {x}^{2} + 1)}$

Squaring on both sides

$\rm \Rightarrow {4}^{2} = (\sqrt{7{x}^{2} + 7})^{2}$

$\rm \Rightarrow 16 = 7{x}^{2} + 7$

$\rm \Rightarrow 16 - 7= 7{x}^{2}$

$\rm \Rightarrow 9= 7{x}^{2}$

$\rm \Rightarrow {x}^{2} = \dfrac{9}{7}$

Taking square root on both sides

$\rm \Rightarrow x = \pm\dfrac{3}{ \sqrt{7} }$