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Sujit Kumar
111 Points
2 years ago
QUESTION 2:

$\frac{1}{sec\theta-tan\theta}-\frac{1}{cos\theta}=\frac{1}{cos\theta}-\frac{1}{sec\theta+tan\theta}$

LHS:
$\frac{1}{sec\theta-tan\theta}-\frac{1}{cos\theta}$
$\rightarrow \frac{sec\theta+tan\theta }{(sec\theta-tan\theta)(sec\theta+tan\theta)}-\frac{1}{cos\theta}$
$\rightarrow \frac{sec\theta+tan\theta }{sec^{2}\theta-tan^{2}\theta}-\frac{1}{cos\theta}$

WE KNOW:
$sec^{2}\theta=1+tan^{2}\theta$

$\rightarrow sec\theta+tan\theta-\frac{1}{cos\theta}$
$\rightarrow sec\theta+tan\theta-sec\theta$
$\rightarrow tan\theta$

RHS:
$\frac{1}{cos\theta}-\frac{1}{sec\theta+tan\theta}$
$\rightarrow \frac{1}{cos\theta}-\frac{sec\theta-tan\theta }{(sec\theta+tan\theta)(sec\theta-tan\theta)}$
$\rightarrow \frac{1}{cos\theta}-\frac{sec\theta-tan\theta }{sec^{2}\theta-tan^{2}\theta}$

WE KNOW:
$sec^{2}\theta=1+tan^{2}\theta$

$\rightarrow \frac{1}{cos\theta}-(sec\theta-tan\theta)$
$\rightarrow sec\theta-sec\theta+tan\theta$
$\rightarrow tan\theta$

LHS=RHS

Hence Proved! Hence Proved! Hence Proved! Hence Proved!
Sujit Kumar
111 Points
2 years ago
QUESTION 3:
$f(x)=\frac{x+3}{4x-5}$      $g(x)=\frac{3x+5}{4x-1}$
$(fog)(x)=f(g(x))$
$\rightarrow f(g(x))=\frac{\frac{3+5x}{4x-1}+3}{4(\frac{3+5x}{4x-1})-5}$
ON SIMPLIFICATION WE GET
$\rightarrow f(g(x))=\frac{3+5x+12x-3}{12+20x-20x+5}$
$\rightarrow f(g(x))=x$
Hope You understood!
Hope You understood!
Hope You understood!
Hope You understood!