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Answer 1

$\large\bf\underline{Question:-}$

If x- tan²60° + 3sin²60°-3/4tan²60°= 6 ,find the Value of 'x'.

$\large\bf\underline{Given:-}$

• x- tan²60° + 3sin²60°-3/4tan²60°=6

$\large\bf\underline {To \: find:-}$

• Value of x

$\huge\bf\underline{Solution:-}$

$\small \longmapsto \rm \: x - {tan}^{2} 60 \degree + 3 {sin}^{2} 60 \degree - \frac{3}{4} {tan}^{2} 60 \degree = 6 ......(i)$

We know that,

$\leadsto \bf \: tan60 \degree = \sqrt{3} \\ \\ \leadsto \bf \: sin60 \degree = \frac{ \sqrt{3} }{2}$

Putting Values of tan 60 and sin 60 in the eq.(i).

$\longmapsto \rm \: x - ( \sqrt{3} ) {}^{2} + 3( \frac{ \sqrt{3} }{2} ) {}^{2} - \frac{3}{4} (\sqrt{3} ) {}^{2} = 6 \\ \\ \longmapsto \rm \: x - 3 + 3 \times \frac{3}{4} - \frac{3}{4} \times 3 = 6 \\ \\ \longmapsto \rm \: x - 3 + \frac{9}{4} - \frac{9}{4} = 6 \\ \\ \longmapsto \rm \: x - 3 + \frac{9}{4} - \frac{9}{4} - 6 = 0 \\ \\ \rm \: taking \: LCM\\ \\ \longmapsto \rm \: \frac{4x - 12 + 9 - 9 - 24}{9} = 0 \\ \\ \longmapsto \rm \: 4x - 12 - 24 = 0 \times 4 \\ \\ \longmapsto \rm \: 4x - 36 = 0 \\ \\ \longmapsto \rm \: 4x = 36 \\ \\\longmapsto \rm \: x = \cancel \frac{36}{4} \\ \\ \longmapsto \boxed{\bf\:x = 9} \:$

So, value of x = 9

Answer 2

Question :–

$\: \: { \bold{ If \: \: x - { \tan }^{2} (60^{ \circ} )+ 3 \sin {}^{2} (60^{ \circ} ) - \dfrac{3}{4} { \tan }^{2} (60^{ \circ} ) = 6}} \: \:$ , then find the value of 'x' .

ANSWER :–

GIVEN :–

$\\ \: \: \longrightarrow \: \: { \bold{x - { \tan }^{2} (60^{ \circ} )+ 3 \sin {}^{2} (60^{ \circ} ) - \dfrac{3}{4} { \tan }^{2} (60^{ \circ} ) = 6}} \: \:$

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

$\\ \implies { \bold{x - { \tan }^{2} (60^{ \circ} )+ 3 \sin {}^{2} (60^{ \circ} ) - \dfrac{3}{4} { \tan }^{2} (60^{ \circ} ) = 6}} \: \:$

$\\ \implies { \bold{x + 3 \sin {}^{2} (60^{ \circ} ) - { \tan }^{2} (60^{ \circ} )- \dfrac{3}{4} { \tan }^{2} (60^{ \circ} ) = 6}} \: \:$

$\\ \implies { \bold{x + 3 \sin {}^{2} (60^{ \circ} ) - \dfrac{7}{4} { \tan }^{2} (60^{ \circ} ) = 6}} \: \:$

• We know that –

$\\ \: \: \longrightarrow \: \: { \bold{ \sin (60^{ \circ} )= \dfrac{ \sqrt{3} }{2} }} \: \:$

$\\ \: \: \longrightarrow \: \: { \bold{ \tan (60^{ \circ} )= { \sqrt{3} } }} \: \:$

• So that –

$\\ \implies { \bold{x + 3 { \left(\dfrac{ \sqrt{3} }{2} \right)}^{2} - \dfrac{7}{4} { \left({ \sqrt{3} } \right)}^{2} = 6}} \: \:$

$\\ \implies { \bold{x + 3 { \left(\dfrac{ {3} }{4} \right)}^{} - \dfrac{7}{4} { \left({ {3} } \right)}^{} = 6}} \: \:$

$\\ \implies { \bold{x + { \left(\dfrac{ {9} }{4} \right)}^{} - { \left({ { \frac{21}{4} } } \right)} = 6}} \: \:$

$\\ \implies { \bold{x + { \left(\dfrac{ {9 - 21} }{4} \right)}^{} = 6}} \: \:$

$\\ \implies { \bold{x - { \cancel \dfrac{ {12} }{4} } = 6}} \: \:$

$\\ \implies { \bold{x - 3 = 6}} \: \:$

$\\ \implies { \bold{x = 6 + 3}} \: \:$

$\\ \: \: \longrightarrow \large \: \: { \boxed{ \bold{x = 9}}} \: \:$

$\\ \rule{220}{2}$