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Question : 22 of 160
Marks:
+1,
-0
Solution:
Observe that
tan(‌)=tan(π−‌)=−tan(‌) Now, using
tan(2θ)=‌Let
θ=‌So,
tan(2⋅‌)=‌⇒tan‌=‌ On putting
tan‌=x,
‌1=‌‌⇒‌‌1−x2−2x=0⇒x2+2x−1=0‌⇒‌‌x=−1−√2,√2−1‌‌ But, ‌tan‌>0‌ so, only take ‌tan‌=√2−1‌‌ Thus, ‌tan(‌)=−tan(‌)=−(√2−1)=1−√2‌
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