જો sin alpha = frac{2pq}{p^2 + q^2} હોય,તો se | vedclass

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

Question

(A) આપેલ છે કે $\sin \alpha = \frac{2pq}{p^2 + q^2}$.આપણે જાણીએ છીએ કે $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{4p^2q^2}{(p^2 + q^2)^2} = \frac{

Vedclass · Accepted Answer

$\frac{p-q}{p+q}$

Vedclass · Answer

(A) આપેલ છે કે $\sin \alpha = \frac{2pq}{p^2 + q^2}$.આપણે જાણીએ છીએ કે $\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{4p^2q^2}{(p^2 + q^2)^2} = \frac{(p^2 + q^2)^2 - 4p^2q^2}{(p^2 + q^2)^2} = \frac{(p^2 - q^2)^2}{(p^2 + q^2)^2}$.તેથી,$\cos \alpha = \frac{p^2 - q^2}{p^2 + q^2}$.હવે,$\sec \alpha = \frac{1}{\cos \alpha} = \frac{p^2 + q^2}{p^2 - q^2}$ અને $	an \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{2pq}{p^2 - q^2}$.હવે,$\sec \alpha - 	an \alpha = \frac{p^2 + q^2}{p^2 - q^2} - \frac{2pq}{p^2 - q^2} = \frac{p^2 + q^2 - 2pq}{p^2 - q^2} = \frac{(p-q)^2}{(p-q)(p+q)} = \frac{p-q}{p+q}$.

જો $\sin \alpha = \frac{2pq}{p^2 + q^2}$ હોય,તો $\sec \alpha - \tan \alpha = $

Similar Questions

જો $A, B, C, D$ એ ચક્રીય ચતુષ્કોણના ખૂણાઓ હોય,તો $\cos A + \cos B + \cos C + \cos D = $

જો $\sec \theta - \tan \theta = \frac{a+1}{a-1}$ હોય,તો $\cos \theta =$

જો $\tan \theta = \frac{a}{b}$ હોય,તો $\frac{{\sin \theta }}{{{{\cos }^8}\theta }} + \frac{{\cos \theta }}{{{{\sin }^8}\theta }} = $

$6(\sin^6 \theta + \cos^6 \theta) - 9(\sin^4 \theta + \cos^4 \theta) + 4$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module