Question

Assertion :STATEMENT 1: ${\tan }^{-1}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)=\frac{\pi }{4}$ Reason: STATEMENT 2: For $x>0,y>0,{\tan }^{-1}\left(\frac{x}{y}\right)+{\tan }^{-1}\left(\frac{y-x}{y+x}\right)=\frac{\pi }{4}$

• A. Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1
• B. Both the statements are TRUE and STATEMENT 2 is NOT the correct explanation of STATEMENT 1
• C. STATEMENT 1 is TRUE and STATEMENT 2 is FALSE
• D. STATEMENT 1 is FALSE and STATEMENT 2 is TRUE

Answer 1

The correct option is A Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1

$ta{n}^{-1}\left(\frac{x}{y}\right)+ta{n}^{-1}\left(\frac{y-x}{y+x}\right)$

$=ta{n}^{-1}\left(\frac{x}{y}\right)+ta{n}^{-1}\left(\frac{1-\frac{x}{y}}{1+\frac{x}{y}}\right)$

$=ta{n}^{-1}\left(\frac{x}{y}\right)+ta{n}^{-1}\left(1\right)-ta{n}^{-1}\left(\frac{x}{y}\right)$

$=ta{n}^{-1}\left(1\right)$

$=\frac{\pi }{4}$.

Similarly

$ta{n}^{-1}\left(\frac{3}{4}\right)+ta{n}^{-1}\left(\frac{1}{7}\right)$

$=ta{n}^{-1}\left(\frac{3}{4}\right)+ta{n}^{-1}\left(\frac{4-3}{4+3}\right)$

$=ta{n}^{-1}\left(\frac{3}{4}\right)+ta{n}^{-1}\left(\frac{1-\frac{3}{4}}{1+\frac{3}{4}}\right)$

$=ta{n}^{-1}\left(\frac{3}{4}\right)+ta{n}^{-1}\left(1\right)-ta{n}^{-1}\left(\frac{3}{4}\right)$

$=ta{n}^{-1}\left(1\right)$
$=\frac{\pi }{4}$

Thus the reason is the correct explanation for the assertion.