Question

If x sin 45º tan 60°=
sin 30° cot 30°/
3 cos 60° cosec 45°
find x.​

Answer 1

Answer:

x sin45° tan60° = sin30° cot30°/ 3cos60° cosec45°

where x=?

Your Question?

My Answer.

x*1/√2*√3= 1/2*√3/3*1/2*√2

x*√3/√2 = √3/2/3/2*√2

x*√3/√2 = √3/2/3√2/2

Know we multiply right hand side.

x*√3/√2 = √3*2/2**3√2

x*√3/√2 = 2√3/6√2

x = 2√3*√2/6√2*√3

we will cancel out √3 and √3, √2 and √2, and we will cut 6 by 2.

know,

x = 1/3.

Here's your Answer.

Please recheck your answer by keeping 1/3 in the place of x. Then Your Answer will be right.

Answer 2

GIVEN :–

$\bf \to x \sin( {45}^{ \circ}) \tan( {60}^{ \circ} ) = \dfrac{ \sin( {30}^{ \circ} ) \cot( {30}^{ \circ} )}{3 \cos( {60}^{ \circ}) cosec( {45}^{ \circ} ) }$

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

$\bf \implies x \sin( {45}^{ \circ}) \tan( {60}^{ \circ} ) = \dfrac{ \sin( {30}^{ \circ} ) \cot( {30}^{ \circ} )}{3 \cos( {60}^{ \circ}) cosec( {45}^{ \circ} ) }$

• We know that –

$\bf \: \: \: \: { \huge{.}} \: \: \sin( {45}^{ \circ}) = \dfrac{1}{ \sqrt{2} }$

$\bf \: \: \: \: { \huge{.}} \: \: \tan( {60}^{ \circ}) = \sqrt{3}$

$\bf \: \: \: \: { \huge{.}} \: \: \sin( {30}^{ \circ}) = \dfrac{1}{2}$

$\bf \: \: \: \: { \huge{.}} \: \: \cot( {30}^{ \circ}) = \sqrt{3}$

$\bf \: \: \: \: { \huge{.}} \: \: \cos( {60}^{ \circ}) = \dfrac{1}{2}$

$\bf \: \: \: \: { \huge{.}} \: \: cosec({45}^{ \circ}) = \sqrt{2}$

• Now put the values –

$\bf \implies x \left(\dfrac{1}{ \sqrt{2}} \right)( \sqrt{3} ) = \dfrac{\left( \dfrac{1}{2} \right) ( \sqrt{3} )}{3\left( \dfrac{1}{2} \right) ( \sqrt{2} ) }$

$\bf \implies x \left(\dfrac{1}{ \sqrt{2}} \right)= \dfrac{1}{3( \sqrt{2} ) }$

$\bf \implies \large{ \boxed{ \bf x= \dfrac{1}{3}}}$

$\rule{150}{3}$