A AND B ARE COMPLIMENTARY angle
Prove that
1+tan A/tan B=sec ²A
Answers
Answered by
2
Trigonometry,
We have,
A + B = 90°.....(1)
Have to proof that
1 + tanA/tanB =sec²A
Now,
L.H.S.= 1 + tanA/tanB
= 1 + tanA/tan(90°- A) [ from equation 1]
= 1 + tanA/cotA [ tan(90°- e) = cote]
= 1 + tan²A
= sec²A [ as 1 + tan²e = sec²e ]
= R.H.S.
That's it
Hope it helped (^_^メ)
We have,
A + B = 90°.....(1)
Have to proof that
1 + tanA/tanB =sec²A
Now,
L.H.S.= 1 + tanA/tanB
= 1 + tanA/tan(90°- A) [ from equation 1]
= 1 + tanA/cotA [ tan(90°- e) = cote]
= 1 + tan²A
= sec²A [ as 1 + tan²e = sec²e ]
= R.H.S.
That's it
Hope it helped (^_^メ)
Similar questions