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please please please answer all the questions as i need to prepare for tomorrow’s exam....please answer all the question...please..please..please.. thanks in advance

please please please answer all the questions as i need to prepare for tomorrow’s exam....please answer all the question...please..please..please..
thanks in advance

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Grade:11

2 Answers

Sujit Kumar
111 Points
5 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
5 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!

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