What is the simplified value of \(\frac{\cos(2\pi-x)}{\sin(\pi+x)}\)?\(\frac{\cos(2\pi-x)}{\sin(\pi+x)}\) का सरल मान क्या है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(-\cot x\)\(-\cot x\)
Using the identities \(\cos(2\pi-x)=\cos x\) and \(\sin(\pi+x)=-\sin x\),
\[\frac{\cos(2\pi-x)}{\sin(\pi+x)}=\frac{\cos x}{-\sin x}=-\cot x.\]
Therefore, the correct answer is \(-\cot x\). Choosing \(\cot x\) would ignore the negative sign in the denominator. Exam tip: for an angle of the form \(\pi+x\), sine has a negative sign.सर्वसमिकाओं \(\cos(2\pi-x)=\cos x\) तथा \(\sin(\pi+x)=-\sin x\) का प्रयोग करने पर
\[\frac{\cos(2\pi-x)}{\sin(\pi+x)}=\frac{\cos x}{-\sin x}=-\cot x.\]
अतः सही उत्तर \(-\cot x\) है। \(\cot x\) लेने पर हर के ऋण चिह्न की उपेक्षा हो जाएगी। परीक्षा टिप: \(\pi+x\) वाले कोण में sine का चिह्न ऋणात्मक होता है।
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
