Question

Plz solve the following:

Answer 1

Note: I am writing theta as A because It is difficult for me to write theta always.


LHS:

Given cosec A - sin A = a^3

$\frac{1}{sinA} - sinA = a^3$

$\frac{1 -sin^2A}{sinA} = a^3$

$\frac{cos^2A}{sinA} = a^3$

$\frac{ cos^{ \frac{2}{3} } A}{ sin^{ \frac{1}{3} }A } = a$

$\frac{ cos^{ \frac{4}{3} }A }{ sin^{ \frac{2}{3} }A } = a^2 ------- (1)$



Given secA - cosb = b^3.

$\frac{1}{cosA} - cosA = b^3$

$\frac{1-cos^2A}{cosA} = b^3$

$\frac{sin^2A}{cosA} = b^3$

$\frac{sin^ \frac{4}{3} A}{cos^ \frac{2}{3} A} = b^2 --------- (2)$


Now,

$a^2 * b^2 = \frac{cos^ \frac{4}{3} A}{sin^ \frac{2}{3} A} * \frac{sin^ \frac{4}{3} A }{cos^ \frac{2}{3} A}$

                      $= sin^ \frac{2}{3} A cos^ \frac{2}{3} A.$




$a^2 + b^2 = \frac{cos^ \frac{4}{3}A }{sin^ \frac{2}{3}A } + \frac{sin^ \frac{4}{3} A}{cos^ \frac{2}{3} A}$

                        $= \frac{cos^2A + sin^2A}{sin^ \frac{2}{3} A cos^ \frac{2}{3}A }$

                        $= \frac{1}{sin^ \frac{2}{3} A cos^ \frac{2}{3}A }$.



Now,

RHS:


$a^2b^2(a^2 + b^2) = sin^ \frac{2}{3} A * cos^ \frac{2}{3} A * \frac{1}{sin^ \frac{2}{3} A * cos \frac{2}{3} A}$


                                       = 1.




LHS = RHS.


Hope this helps!