If sin theta + cos theta = m and sec theta + | vedclass

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

Question

(C) Given: $\sin 	heta + \cos 	heta = m$ and $\sec 	heta + 	ext{cosec} 	heta = n$. Squaring the first equation: $(\sin 	heta + \cos 	heta)^2 =

Vedclass · Accepted Answer

$2m$

Vedclass · Answer

(C) Given: $\sin 	heta + \cos 	heta = m$ and $\sec 	heta + 	ext{cosec} 	heta = n$. Squaring the first equation: $(\sin 	heta + \cos 	heta)^2 = m^2 \implies 1 + 2 \sin 	heta \cos 	heta = m^2 \implies 2 \sin 	heta \cos 	heta = m^2 - 1$. Now,consider the expression $n(m + 1)(m - 1) = n(m^2 - 1)$. Substitute $n = \sec 	heta + 	ext{cosec} 	heta = \frac{1}{\cos 	heta} + \frac{1}{\sin 	heta} = \frac{\sin 	heta + \cos 	heta}{\sin 	heta \cos 	heta} = \frac{m}{\sin 	heta \cos 	heta}$. Thus,$n(m^2 - 1) = \left( \frac{m}{\sin 	heta \cos 	heta} ight) (2 \sin 	heta \cos 	heta) = 2m$.

If $\sin \theta + \cos \theta = m$ and $\sec \theta + \text{cosec} \theta = n$,then $n(m + 1)(m - 1) = $

Similar Questions

One root of the equation $\cos x - x + \frac{1}{2} = 0$ lies in the interval

The number of all possible integral values of $n > 2$ such that $\sin \frac{\pi}{2n} + \cos \frac{\pi}{2n} = \frac{\sqrt{n}}{2}$ is:

$4 \cos \frac{\pi}{7} \cos \frac{\pi}{5} \cos \frac{2 \pi}{7} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{7} = $

If a regular pentagon and a regular decagon have the same perimeter,then the ratio of their areas is

If $\cosh \alpha + \sinh \alpha = e^3$ and $\sinh x = \frac{\alpha}{\alpha+1}$,then $\tanh x =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module