frac{1-sin theta+cos theta}{1-sin theta-cos t | vedclass

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

Question

(B) આપણે નિત્યસમ $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ અને $\cos 	heta = 2 \cos^2 \frac{	heta}{2} - 1 = 1 - 2 \sin^2 \frac{\

Vedclass · Accepted Answer

$-\cot \frac{	heta}{2}$

Vedclass · Answer

(B) આપણે નિત્યસમ $\sin 	heta = 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}$ અને $\cos 	heta = 2 \cos^2 \frac{	heta}{2} - 1 = 1 - 2 \sin^2 \frac{	heta}{2}$ નો ઉપયોગ કરીએ છીએ.આ કિંમતો પદાવલિમાં મૂકતા:$\frac{1 - (2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}) + (2 \cos^2 \frac{	heta}{2} - 1)}{1 - (2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}) - (1 - 2 \sin^2 \frac{	heta}{2})}$$= \frac{2 \cos^2 \frac{	heta}{2} - 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}{2 \sin^2 \frac{	heta}{2} - 2 \sin \frac{	heta}{2} \cos \frac{	heta}{2}}$$= \frac{2 \cos \frac{	heta}{2} (\cos \frac{	heta}{2} - \sin \frac{	heta}{2})}{2 \sin \frac{	heta}{2} (\sin \frac{	heta}{2} - \cos \frac{	heta}{2})}$$= \frac{2 \cos \frac{	heta}{2} (\cos \frac{	heta}{2} - \sin \frac{	heta}{2})}{-2 \sin \frac{	heta}{2} (\cos \frac{	heta}{2} - \sin \frac{	heta}{2})}$$= -\frac{\cos \frac{	heta}{2}}{\sin \frac{	heta}{2}} = -\cot \frac{	heta}{2}$.

$\frac{1-\sin \theta+\cos \theta}{1-\sin \theta-\cos \theta} = $

Similar Questions

$3 \tan^6 10^{\circ} - 27 \tan^4 10^{\circ} + 33 \tan^2 10^{\circ} = $

$\tan \alpha + 2 \tan 2 \alpha + 4 \tan 4 \alpha + 8 \cot 8 \alpha = $

જો $a \tan \theta = b$ હોય,તો $a \cos 2\theta + b \sin 2\theta = $

જો $\tan x = -\frac{4}{3}$ અને $x$ એ બીજા ચરણમાં હોય,તો $\sin \frac{x}{2}, \cos \frac{x}{2}$ અને $\tan \frac{x}{2}$ શોધો.

જો $\sin(x+3\alpha) + 3\sin(x-\alpha) = 0$ હોય,તો

Vedclass Test Series

Exam Paper Generator

Online Exam Module