Question

plz answer this question guys​

Answer 1

Answer:

GIVES THAT :–

$\tan( \alpha + \beta ) = \sqrt{3}$

And —

$\tan( \alpha - \beta ) = 1$

SOLVING THESE EQUATION AND GET –

$\alpha + \beta = {60}^{0} \\ \\ \alpha - \beta = {45}^{0}$

SO ,

$\alpha = {52.5}^{0}$

$\beta = {7.5}^{0}$

SO THAT –

$= \tan(10 \alpha ) \\ \\ = \tan(10 \times {52.5}^{0} ) \\ \\ = \tan( {525}^{0} ) \\ \\ = \tan( {360}^{0} + {165}^{0} ) \\ \\ = \tan( {165}^{0} ) \\ \\ = \tan(180 {}^{0} - 15 {}^{0} ) \\ \\ = - \tan(15 {}^{0} )$

WE KNOW THAT –

$\tan(15 {}^{0} ) = 2 - \sqrt{3}$

So,

$= - (2 - \sqrt{3} ) \\ \\ = \sqrt{3} - 2$

☆☆☆ OPTION (4) IS CORRECT ☆☆☆

☆☆☆ FOLLOW ME ☆☆☆

Answer 2

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

Actually Welcome to the Concept of the Trigonometry

Now here basically we are going to shift the trigonometry, and convert it into the Inverse trigonometry.

for this, we will first do

alpha + beta = tan^-1(root 3 )

and from the other equation we get as,

alpha - beta = tan^-1(1)

so after solving we get as,

alpha + beta = π/3

alpha - beta = π/4

adding both the equation we get as,

2 alpha = 7π/12

now multiplying by "5" in both sides we get as,

10 alpha= 35π/12

taking tan ratio on both sides, we get as,

tan (10 alpha) = tan(35π/12)

tan(10 alpha) = tan ( 5π/3+5π/4)

now applying the formula as

⭐tan ( A+B) = tanA + tanB / 1 - tanA.tanB