Question

Prove the identity​

Answer 1

HOPE THIS WILL HELP YOU

Answer 2

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