Mathematics Answer, Explanation and Revision Hints
यदि \(\cos^2 x=\frac{9}{16}\) और (x) चौथे चतुर्थांश में है, तो \(\sin x\) का मान क्या है? / If \(\cos^2 x=\frac{9}{16}\) and (x) is in the fourth quadrant, what is the value of \(\sin x\)?
Correct Answer: B. \(-\frac{\sqrt{7}}{4}\). Explanation: \(\sin^2 x=1-\frac{9}{16}=\frac{7}{16}\) होता है। चौथे चतुर्थांश में \(\sin x\) ऋणात्मक होता है। / \(\sin^2 x=1-\frac{9}{16}=\frac{7}{16}\). In the fourth quadrant, \(\sin x\) is negative.
Which concept should I revise for this Mathematics MCQ?
\(\sin^2 x=1-\frac{9}{16}=\frac{7}{16}\). In the fourth quadrant, \(\sin x\) is negative.
What exam hint can help solve this Mathematics question?
\(\sin^2 x=1-\frac{9}{16}=\frac{7}{16}\) होता है। चौथे चतुर्थांश में \(\sin x\) ऋणात्मक होता है।
Login to view answers
Use Google login or mobile OTP. Admin can block users and control login providers.
Student Class Required
Select your class first
Quiz questions, daily challenge and practice pages will open according to your selected class. Class 11/12 ke liye stream bhi select karein.
