Question

Question number 103 no spamming

Answer 1

the answer should be the second one because se it is

pie by 2 - and in second bracket there is + so

Answer 2

Answer:

$f(x) = \frac{tan x}{\sqrt{1 + tan^{2}x}}$

$sec^{2}x - tan^{2} = 1\\\\sec^{2}x = 1 + tan^{2}x$

$f(x) = \frac{tan x}{\sqrt{sec^{2}x }}\\\\f(x) = \frac{tanx}{secx}\\\\f(x) = \frac{\frac{sinx}{cosx}}{\frac{1}{cosx}}\\\\f(x) = sinx$

now

$x = \frac{- \pi }{2}$

$f(x) = a\\\\sin(\frac{-\pi }{2}) = a\\\\sin(\frac{-22}{7 * 2} ) = a\\\\sin(\frac{-11}{7}) = a\\\\sin = \frac{-7a}{11}$

& also

$x = \frac{ \pi }{2}$

$f(x) = b\\\\sin(\frac{\pi }{2}) = b\\\\sin(\frac{22}{7 * 2} ) = b\\\\sin(\frac{11}{7}) = b\\\\sin = \frac{7b}{11}$

from the above we conclude that

$\frac{-7a}{11} = \frac{7b}{11}\\\\-a = b$

I hope it is correct & it helps..........