Differentiation Question 90
Question: If $ \tan (x+y)+\tan (x-y)=1, $ then $ \frac{dy}{dx}= $
[DSSE 1979]
Options:
A) $ \frac{{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)}{{{\sec }^{2}}(x+y)-{{\sec }^{2}}(x-y)} $
B) $ \frac{{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)}{{{\sec }^{2}}(x-y)-{{\sec }^{2}}(x+y)} $
C) $ \frac{{{\sec }^{2}}(x+y)-{{\sec }^{2}}(x-y)}{{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ \tan (x+y)+\tan (x-y)=1 $
Differentiating w.r.t. x of y, we get
Therefore $ {{\sec }^{2}}(x+y)( 1+\frac{dy}{dx} )+{{\sec }^{2}}(x-y)( 1-\frac{dy}{dx} )=0 $
Therefore $ \frac{dy}{dx}=\frac{{{\sec }^{2}}(x+y)+{{\sec }^{2}}(x-y)}{{{\sec }^{2}}(x-y)-{{\sec }^{2}}(x+y)} $ .
BETA
