Question

Assertion :If $sec\alpha ,cosec\alpha$ are the roots of the equation $x^2-px+q=0$, then $p^2-2q-q^2=0$. Reason: If p, q are the roots of the equation $x^2-xsin\alpha +cos\alpha =0$, then $p^2=\frac{1-q^2}{1+q^2}$.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

Assertion.
$sec\alpha +cosec\alpha =p$ and $sec\alpha .cosec\alpha =q$

$\Rightarrow \frac{1}{cos\alpha }+\frac{1}{sin\alpha }=p$ and $\frac{1}{(cos\alpha .sin\alpha )}=q$

$\Rightarrow cos\alpha +sin\alpha =p.cos\alpha .sin\alpha$ and $q.cos\alpha .sin\alpha =1$

$\Rightarrow (cos\alpha +sin\alpha {)}^2=(p.cos\alpha .sin\alpha {)}^2$

$\Rightarrow 1+2cos\alpha .sin\alpha =p^2.co{s}^2\alpha .si{n}^2\alpha$

$\Rightarrow 1+\frac{2}{q}=\frac{p^2}{q^2}\Rightarrow p^2-2q-q^2=0$

Hence Assertion is correct.

Reason.

$p+q=sin\alpha$ and $p.q=cos\alpha$

squaring and adding,

$p^2+q^2+2p.q+p^2.q^2=1$

$\Rightarrow p^2=\frac{1-q^2(p^2+1)}{p+2q}$

Hence Reason is incorrect.

thus correct choice is 'C'.