Question

plz help me to find this solution​

Answer 1

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

• Evaluate the expression
• $\sf \to \: \: \: \frac{1 - { \tan}^{2} 30 \degree }{ 1 + { \cot}^{2}30 \degree }$

${ \large{ \natural }}\: \: \: \sf \: Concept \: \: required \: \: to \: \: solve \: \: this \: \: problem,$

• We need to know the exact value of tan( ratio of high and base ) and cot( ratio of base and hight ) at 30 degree first.
• Secondly it's definite accordingly with angles.

$\bf \large \maltese \: \: \: { \underline{\underline{Solution : }}}$

$\: \tt \underline{ \underline{ {1}^{st} \: \: Step : }}$

$\sf \: Lets \: find \: out \: the \: value \: of \tan \: and \cot \: at \: 30 \degree$

Lets assume ∆ABC is a Right triangle having length of each sides a

As it's a Right triangle then angles will be 60° [ property of Right triangle ]

now we drew a right angle on BC from A ;

hence, DC = a/2

hence, tan 60° = AD/DC

Now, We need to find out the length of AD to evaluate the value of tan 60°

We know sin is the ratio of Hight and hypotenuse

hence, sin 60° = Hight/Hypotenuse = AD/a

=> a sin 60° = AD ____( 1 )

From Pythagoras theorem,

base²+hight² = hypotenuse ²

=> (a/2)² + AD² = a²

=> AD² = a² - a²/4

=> AD² = 3a/4

=> AD = √3a/2

substituting value in eqn ( 1 )

a sin 60° = √3a/2

sin 60° = √3/2____( 2 )

again cos 60° = base/hypotenuse

cos 60° = (a/2)/a => 1/2____( 3 )

as we know sin a / cos a = tan a

hence we can say tan 60 = ( √3/2 )/( 1/2 ) => √3

again we know tan a is inverse of cot a

hence cot 60° = 1/√3

$\: \tt \underline{ \underline{ {2}^{nd} \: \: Step : }}$

$\sf \: Lets \: \: substitute \: \: the \: \: values \: \: we \: \: have \: \: got$

$\: \: \: \frac{1 - { \tan}^{2} \theta }{1 + { \cot}^{2} \theta}$

$\implies \frac{1 - { \big( \sqrt{3} \big)}^{2} }{1 + { \big( \frac{1 }{ \sqrt{3} } \big) }^{2} }$

$\implies \frac{1 - 3}{1 - \frac{1}{3} }$

$\implies \frac{ - \cancel{2}}{ \frac{ \cancel{2}}{3} }$

$\implies \bf - 3$