જો $\cos x = -\frac{1}{2}$ અને $x$ ત્રીજા ચરણમાં હોય,તો બાકીના પાંચ ત્રિકોણમિતીય વિધેયોની કિંમત શોધો.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
આપેલ છે $\cos x = -\frac{1}{2}$.
$\sec x = \frac{1}{\cos x} = \frac{1}{(-1/2)} = -2$.
નિત્યસમ $\sin^2 x + \cos^2 x = 1$ નો ઉપયોગ કરતા:
$\sin^2 x = 1 - \cos^2 x = 1 - (-1/2)^2 = 1 - 1/4 = 3/4$.
$x$ ત્રીજા ચરણમાં હોવાથી,$\sin x$ ઋણ હશે.
$\sin x = -\frac{\sqrt{3}}{2}$.
$\csc x = \frac{1}{\sin x} = -\frac{2}{\sqrt{3}}$.
$\tan x = \frac{\sin x}{\cos x} = \frac{-\sqrt{3}/2}{-1/2} = \sqrt{3}$.
$\cot x = \frac{1}{\tan x} = \frac{1}{\sqrt{3}}$.

Explore More

Similar Questions

$\tan x + \frac{\cos x}{1 + \sin x} = $

$\sin ^4 \frac{\pi}{8} + \cos ^4 \frac{3 \pi}{8} - \sin ^4 \frac{3 \pi}{8} + \sin ^4 \frac{5 \pi}{8} + \cos ^4 \frac{7 \pi}{8} - \sin ^4 \frac{7 \pi}{8} = ?$

$\frac{\sqrt{3} \sin \theta + \cos \theta}{\sin \left(\theta + \frac{\pi}{6}\right)} = $

$\sin 50^\circ - \sin 70^\circ + \sin 10^\circ = $

$\operatorname{cosec} 15^{\circ} + \sec 15^{\circ}$ ની કિંમત શોધો:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo