If (x + y) + (x - y) = 1, then dy dx = - Vidyadip

MCQ

If $\tan (x + y) + \tan (x - y) = 1,$ then ${{dy} \over {dx}} = $

A
${{{{\sec }^2}(x + y) + {{\sec }^2}(x - y)} \over {{{\sec }^2}(x + y) - {{\sec }^2}(x - y)}}$
✓
${{{{\sec }^2}(x + y) + {{\sec }^2}(x - y)} \over {{{\sec }^2}(x - y) - {{\sec }^2}(x + y)}}$
C
${{{{\sec }^2}(x + y) - {{\sec }^2}(x - y)} \over {{{\sec }^2}(x + y) + {{\sec }^2}(x - y)}}$
D
None of these

✓

Answer

Correct option: B.

${{{{\sec }^2}(x + y) + {{\sec }^2}(x - y)} \over {{{\sec }^2}(x - y) - {{\sec }^2}(x + y)}}$

b
(b) $\tan (x + y) + \tan (x - y) = 1$

Differentiating w.r.t. $x$ of $y,$ we get

==> ${\sec ^2}(x + y)\left( {1 + \frac{{dy}}{{dx}}} \right) + {\sec ^2}(x - y)\left( {1 - \frac{{dy}}{{dx}}} \right) = 0$

==> $\frac{{dy}}{{dx}} = \frac{{{{\sec }^2}(x + y) + {{\sec }^2}(x - y)}}{{{{\sec }^2}(x - y) - {{\sec }^2}(x + y)}}$.

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