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If `tan A` and `tan B` are the roots of `x^2-px + q = 0` , then the value of` sin^2(A+ B)` isA. `(p^(2))/(p^(2)+(1+q)^(2))`B. `(p^(2))/(p^(2)+q^(2))`C. `(q^(2))/(p^(2)+(1-q)^(2))`D. `(p^(2))/((p+q)^(2))` |
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Answer» Correct Answer - A We have, `tan A + tan B = p and tan A tan B = q`. `therefore" "tan(A+B) = (tan A + tan B)/(1-tan A tan B)=(p)/(1-q)` Now, `sin^(2)(A+B) = (1)/(2){1 - cos 2 (A+B)}` `rArr" "sin^(2)(A+B)=(1)/(2){1-(2-tan^(2)(A+B))/(1+tan^(2)(A+B))}=(tan^(2)(A+B))/(1+tan^(2)(A+B))` `rArr" "sin^(2)(A+B)=(p^(2)//(1-q)^(2))/(1+(p^(2))/(1-q)^(2))=(p^(2))/(p^(2)+(1-q)^(2))` |
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