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

The value of $\cos y\cos\big(\large\frac{\pi}{2}$$-x)$$-\cos(\large\frac{\pi}{2}$$-y\big)\cos x+\sin y\cos\big(\large\frac{\pi}{2}-x\big)$$+\cos x.\sin\big(\large\frac{\pi}{2}$$-y\big)$ is zero.If

$(a)\;x=0\qquad(b)\;y=0\qquad(c)\;x=y\qquad(d)\;x=n\pi-\large\frac{\pi}{4}$$+y$

Can you answer this question?
 
 

1 Answer

0 votes
$\cos y\cos\big(\large\frac{\pi}{2}$$-x\big)-\cos\big(\large\frac{\pi}{2}$$-y\big)\cos x+\sin y\cos\big(\large\frac{\pi}{2}$$-x\big)+\sin y\cos\big(\large\frac{\pi}{2}-$$x\big)+\cos x\sin\big(\large\frac{\pi}{2}$$-y\big)=0$
$\Rightarrow \cos y\sin x-\cos x\sin y+\sin y\sin x+\cos x\cos y=0$
$\Rightarrow \sin(x-y)+\cos(x-y)=0$
$\Rightarrow \tan(x-y)=-1$
$\Rightarrow x=n\pi-\large\frac{\pi}{4}$$+y$
Hence (d) is the correct answer.
answered Oct 9, 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
...