JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 11)
If tanA and tanB are the roots of the quadratic equation, 3x2 $$-$$ 10x $$-$$ 25 = 0, then the value of 3 sin2(A + B) $$-$$ 10 sin(A + B).cos(A + B) $$-$$ 25 cos2(A + B) is :
$$-$$ 10
10
$$-$$ 25
25
Explanation
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
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
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