If sin alpha and cos alpha are the roots of t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given that $\sin \alpha$ and $\cos \alpha$ are roots of $ax^2 + bx + c = 0$.From the properties of quadratic equations,the sum of roots is $\sin \

Vedclass · Accepted Answer

$a^2 - b^2 + 2ac = 0$

Vedclass · Answer

(A) Given that $\sin \alpha$ and $\cos \alpha$ are roots of $ax^2 + bx + c = 0$.From the properties of quadratic equations,the sum of roots is $\sin \alpha + \cos \alpha = -\frac{b}{a}$ and the product of roots is $\sin \alpha \cos \alpha = \frac{c}{a}$.We know the identity $\sin^2 \alpha + \cos^2 \alpha = 1$.This can be written as $(\sin \alpha + \cos \alpha)^2 - 2 \sin \alpha \cos \alpha = 1$.Substituting the values,we get $(-\frac{b}{a})^2 - 2(\frac{c}{a}) = 1$.$\frac{b^2}{a^2} - \frac{2c}{a} = 1$.Multiplying by $a^2$,we get $b^2 - 2ac = a^2$.Rearranging the terms,we get $a^2 - b^2 + 2ac = 0$.

If $\sin \alpha$ and $\cos \alpha$ are the roots of the equation $ax^2 + bx + c = 0$ $(c \neq 0)$,then:

Similar Questions

If $\frac{\cos^4 \alpha}{\cos^2 \beta} + \frac{\sin^4 \alpha}{\sin^2 \beta} = 1$,then the value of $\left[ \frac{\cos^4 \beta}{\cos^2 \alpha} + \frac{\sin^4 \beta}{\sin^2 \alpha} \right]$ is (where $[.]$ denotes the greatest integer function).

If $\sin A = \frac{1}{\sqrt{10}}$ and $\sin B = \frac{1}{\sqrt{5}}$,where $A$ and $B$ are positive and acute angles,then $A + B = $

$\frac{2\sin \theta \tan \theta (1 - \tan \theta ) + 2\sin \theta \sec^2 \theta}{(1 + \tan \theta )^2} = $

$\frac{{\sin (B + A) + \cos (B - A)}}{{\sin (B - A) + \cos (B + A)}} = $

The maximum value of the function $f(x) = 3\sin x + 4\cos x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module