logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
0 votes

If $\large\frac{\sin^4x}{2}+\frac{\cos^4x}{3}=\frac{1}{5}$ then

$\begin{array}{1 1}(a)\;\tan^2x=0&(b)\;\large\frac{\sin^8x}{8}+\frac{\cos^8x}{27}=\frac{1}{125}\\(c)\;\tan^2x=\large\frac{1}{3}&(d)\;\large\frac{\sin^8x}{8}+\frac{\cos^8x}{27}=\frac{2}{125}\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Given :
$\large\frac{\sin^4 x}{2}+\frac{\cos^4x}{3}=\frac{1}{5}$
$\Rightarrow 3\sin^4 x+2\cos^4x=\large\frac{6}{5}$
$\Rightarrow \sin^4 x+2[\sin^4x+\cos^4x]=\large\frac{6}{5}$
$\Rightarrow \sin^4x+2[1-2\sin^2x\cos^2x]=\large\frac{6}{5}$
$\sin^4x+2-4\sin^2x(1-\sin^2x)=\large\frac{6}{5}$
$5\sin^4x-4\sin^2x+2-\large\frac{6}{5}$$=0$
$25\sin^4x-20\sin^2x+4=0$
$\Rightarrow (5\sin^2x-2)^2=0$
$\Rightarrow \sin^2 x=\large\frac{2}{5}$
$\Rightarrow \cos^2x=\large\frac{3}{5}$
$\Rightarrow \tan^2 x=\large\frac{2}{3}$
$\large\frac{\sin^8x}{8}+\frac{\cos^8x}{27}=\frac{2}{625}+\frac{3}{625}$
$\qquad\qquad\qquad=\large\frac{5}{625}$
$\qquad\qquad\qquad=\large\frac{1}{125}$
Hence (b) is the correct answer.
answered Oct 14, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...