Question

☆ content quality required ☆


if
$tan \: a \: tan \: b \: = \sqrt{ \frac{a - b}{a + b} }$

Then, prove that
$(a - b \: \cos(2a)) ((a - b \cos(2b) ) = {a}^{2} b {}^{2}$

Answer 1

HELLO DEAR,

IT SEEM'S QUESTIONS HAVE SOME ERROR,

YOUR QUESTIONS IS------------------> if

$\bold{tanAtanB= \sqrt{ \frac{a - b}{a + b} }}$

Then, prove that

$(a - b cos2A)(a - b cos2B) = {a}^{2} - b^{2}$

GIVEN:-

tanAtanB = √(a - b)/(a + b)

$\bold{\Rightarrow tan^2Atan^2B = (a - b)/(a + b)}$

tan²Atan²B(a + b) = (a - b)-------------( 1 )

now,

$\bold{ (a - bcos2A)*(a - bcos2B) = a^2 - b^2}$

we know:- cos2A = (1 - tan ²A)/(1 + tan²A)

so, $\bold{\Rightarrow [a - b\frac{1 - tan^2A}{1 + tan^2B}]*[a - b\frac{1 - tan^2B}{1 + tan^2B}]}$

$\bold{\Rightarrow [a - \frac{b - btan^2A}{1 + tan^2B}]*[a - \frac{b - btan^2B}{1 + tan^2B}]}$

$\bold{\Rightarrow \frac{a(1 + tan^2A) - b + btan^2A}{(1 + tan^2A)} * \frac{a(1 + tan^2B) - b + btan^2B}{1 + tan^2B}}$

$\bold{\Rightarrow \frac{(a + atan^2A - b + btan^2A)*(a + atan^2B - b + btan^2B)}{(1 + tan^2A)(1 + tan^2B)}}$

$\bold{\Rightarrow [\frac{(a - b) + (a + b)tan^2A}{(1 + tan^2A)}]*[\frac{(a - b) + (a + b)tan^2B}{(1 + tan^2B)}]}$

$\bold{\Rightarrow [\frac{tan^2Atan^2B(a + b) + (a + b)tan^2A}{(1 + tan^2A)}]*[\frac{tan^2Atan^2B(a + b) + (a + b)tan^2B}{(1 + tan^2B)}]}$

from -------------( 1 )

$\bold{\Rightarrow (a + b)(a + b)(\frac{tan^2Atan^2B + tan^2A}{1 + tan^2A}) * (\frac{tan^2Atan^2B + tan^2B}{1 + tan^2B})}$

$\bold{\Rightarrow (a + b)(a + b)[\frac{tan^2A(1 + tan^2B)*tan^2B(1 + tan^2A)}{(1 + tan^2A)(1 + tan^2B)}]}$

$\bold{\Rightarrow (a + b)(a + b)*tan^2Atan^2B}$

$\bold{\Rightarrow \frac{(a + b)(a + b)(a - b)}{(a + b)}}$

From-------------( 1 )

as,[tan²Atan²B = (a - b)/(a + b)]

$\bold{\therefore, a^2 - b^2}$

$\bold{\boxed{HENCE, (a - bcos2A)(a - bcos2B) = (a^2 - b^2)}}$

I HOPE ITS HELP YOU DEAR,

THANKS

Answer 2

❤️Multiple Thanker❤️

❤️Multiple Thanker❤️

❤️Multiple Thanker❤️

La bsdk ke bete areey tum to bade heavy driver ho bhai