Advertisements
Advertisements
प्रश्न
Which trigonometric substitution is suitable for \[x^2-a^2\] and \[\sqrt{x^2-a^2}\]?
विकल्प
\[x=a\cos\theta\]
\[x=a\tan\theta\]
\[x=a\sin\theta\]
\[x=a\sec\theta\]
MCQ
Advertisements
उत्तर
For \[x^2-a^2\] and \[\sqrt{x^2-a^2}\], use \[x=a\sec\theta\]. This matches the trigonometric identity involving \[\sec^2\theta-1\].
shaalaa.com
क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
