1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Find sinx2,cosx2andtanx2 in each of the following:1:-tanx=−43, x in quadrant II2:-cosx=−13, x in quadrant III

Open in App
Solution

## Part 1st given, tanx=−43 ,x in 2nd quadrant We have to find tanx2=?,cosx2=?,sinx2=? Consider , tanx=−43 tan(2.x2)=−43 2tanx21−tan2x2=−43 3×2tanx2=−4(1−tan2x2) 6tanx2=−4(1−tan2x2)=−4+4tan2x2 4tan2x2−6tanx2−4=0 2tan2x2−3tanx2−2=0 By solving this equation, we get tanx2=−12,tanx2=2 We take tanx2=−12 because in 2nd quadrant tanx is negative Hence , tanx2=−12 tan2x2=(−12)2=14 tan2x2+1=14+1 tan2x2+1=54 sec2x2=54 secx2=±√52 cosx2=±√52 Because in 2nd quadrant cosx is negative cosx2=−√52 Now , cos2x2=(−√52)2=54 1−cos2x2=1−54=−14 sinx2=±√−12 Because in 2nd quadrant cosx is negative sinx2=√−12 Part 2nd .Given, cosx=−13 and x is in III quarant.Now, ∵cosA=2cos2A2−1 ∴cos2A2=√cosA+12 ∴cos2x2=√cosx+12= ⎷−13+12=±√13But cosx is negative in III quadrant cosx2=−√13 ∵sinA2=√1−cosA2=  ⎷1−(−13)2But in II quadrant sinx is negative Sinx2=±√23 =−√23 Tanx2=sinx2cosx2=√2

Suggest Corrections
3
Join BYJU'S Learning Program
Related Videos
Multiple and Sub Multiple Angles
MATHEMATICS
Watch in App
Join BYJU'S Learning Program