Prove the identity
Answers
HOPE THIS WILL HELP YOU
Answer:
(sinA+cosA)
Step-by-step explanation:
Given--->
----------
cosA sinA
--------------- + ----------------- =(sinA+cosA)
(1- tanA) (1-cotA)
proof--->
---------
cosA sinA
L•H•S•=-------------- + ----------------
(1-tanA) (1- cotA)
cosA sinA
=-------------------- + ------------------
sinA cosA
(1 - -----------) (1 - ------------)
cosA sinA
cosA sinA
=--------------------- + ----------------------
cosA - sinA sinA - cosA
(---------------------) (---------------------)
cosA sinA
cos²A sin²A
=---------------------- + ---------------------
(cosA-sinA) (sinA-cosA)
taking (-) common from (sinA-cosA)
cos²A sin²A
=------------------------ - ---------------------
(cosA-sinA) (cosA-sinA)
cos²A - sin²A
=------------------------------------------
(cosA - sinA)
we have an identity , a²-b²=(a+b)(a-b)
applying it in numerator
(cosA+sinA) (cosA-sinA)
=-----------------------------------------
(cosA-sinA)
(cosA-sinA) cancel out from numerator and denominator
=(cosA+sinA)=R.H.S.
Other useful identities---->
------------------------------------
1: sin²θ+cos²θ=1
2: 1+tan²θ=sec²θ
3: 1+cot²θ=cosec²θ
4:tanA=sinA/cosA
5:cotA=cosA/sinA
6: cosesA=1/sinA
7: SecA=1/cosA
Hope it helps u
Thanks