Can u comment the formula here..

Wil be useful for all!!

Wil be useful for all!!

0 votes

0 votes

option D)

The equation is $(\cos \alpha -\lambda )^{2}-\sin ^{2}\alpha =0$

which is a quadratic equation in $\lambda$.

Solve using sridhara acharya's formula. You get option D

The equation is $(\cos \alpha -\lambda )^{2}-\sin ^{2}\alpha =0$

which is a quadratic equation in $\lambda$.

Solve using sridhara acharya's formula. You get option D

0 votes

This is the sridhara acharya's formula which all of us had studied in our 10th standard :p

$x=(-b\pm \sqrt{b^{2}-4ac})/(2*a)$

where a,b,c are the constants of the general form of a quadratic equation ax^2+bx+c=0

$x=(-b\pm \sqrt{b^{2}-4ac})/(2*a)$

where a,b,c are the constants of the general form of a quadratic equation ax^2+bx+c=0

0 votes

Let A be the matrix given in the question.

Now,Trace(A)=sum of eigen values of A

or,Trace(A)=cos α + cos α=2*cos α

Now,by adding and subtracting sin α we get:

or,2cos α=(cos α + sin α) + (cos α - sin α)= *λ**1 +* *λ* 2

Therefore, 2 eigen values can be denoted by (cos ∝ ± sin ∝ (d))

**So, Option (d) is the ans.**