QJEEM18Ma4d99: JEE Main 2018 Mathematics - Class 11 JEEM18Ma4d99 Solution: If tanA and tanB are the roots of the quadratic equation, 3x 2…

As tan A and tan B are the roots of 3x² $-$ 10x $-$ 25 = 0,

So, tan(A + B) = ${{\tan A + \tan B} \over {1 - \tan A\tan B}}$

= ${{{{10} \over 3}} \over {1 + {{25} \over 3}}}$ = ${{10/3} \over {28/3}}$ = ${5 \over {14}}$

Now, cos² (A + B) = $-$ 1 + 2 cos² (A + B)

= ${{1 - {{\tan }^2}(A + B)} \over {1 + {{\tan }^2}(A + B)}}\,$   $\Rightarrow$  cos²(A + B) = ${{196} \over {221}}$

$\therefore\,\,\,$ 3sin²(A + B) $-$ 10sin(A + B)cos(A + B) $-$ 25 cos²(A + B)

= cos²(A + B) [ 3tan²(A + B) $-$ 10tan(A + B) $-$ 25]

= ${{75 - 700 - 4900} \over {196}} \times {{196} \over {221}}$

= $-$ ${{5525} \over {196}}$ $\times$ ${{196} \over {221}}$ = $-$ 25