Answer:
This is the solving step
Step-by-step explanation:
L.H.S.= tan theta+sec theta-1 tan theta-sec theta+1
= (tan theta+sec theta)-(scc^ 2 theta-tan^ 2 theta) (tan theta-sec theta+1)
(**1=sec^ 2 theta-tan^ 2 theta)
= (sec theta+tan theta)[1-(sec theta-tan theta)] (tan theta-sec theta+1)
= (sec theta+tan theta)(tan theta-sec theta+1) (tan theta-sec theta+1)
=sec theta+tan theta = 1 cos theta + sin theta cos theta = 1+sin theta cos theta =R.H.S.
when using 180°of all thetas
tan(theta)+sec(theta)-1/tan(theta)-sec(theta)+1=1+sin(theta)/cos(theta)
0+(-1)-1/0-(-1)+1=1+0/-1
-1-1/+1+1=1/-1
-2/2=1/-1
-1=-1
hence proved,
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Verified answer
Answer:
This is the solving step
Step-by-step explanation:
L.H.S.= tan theta+sec theta-1 tan theta-sec theta+1
= (tan theta+sec theta)-(scc^ 2 theta-tan^ 2 theta) (tan theta-sec theta+1)
(**1=sec^ 2 theta-tan^ 2 theta)
= (sec theta+tan theta)[1-(sec theta-tan theta)] (tan theta-sec theta+1)
= (sec theta+tan theta)(tan theta-sec theta+1) (tan theta-sec theta+1)
=sec theta+tan theta = 1 cos theta + sin theta cos theta = 1+sin theta cos theta =R.H.S.
Step-by-step explanation:
when using 180°of all thetas
tan(theta)+sec(theta)-1/tan(theta)-sec(theta)+1=1+sin(theta)/cos(theta)
0+(-1)-1/0-(-1)+1=1+0/-1
-1-1/+1+1=1/-1
-2/2=1/-1
-1=-1
hence proved,