Question

cosA/1-tanA+sinA/1-cotA = sinA+cosA​

Answer 1

Answer:

Step-by-step explanation:

1−tanA

cosA

​  

+  

1−cotA

sinA

​  

 

1−  

cosA

sinA

​  

 

cosA

​  

+  

1−  

sinA

cosA

​  

 

sinA

​  

 

cosA−sinA

cos  

2

A

​  

−  

cosA−sinA

sin  

2

A

​  

 

cosA−sinA

cos  

2

A−sin  

2

A

​  

 

cosA−sinA

(cosA+sinA)(cosA−sinA)

​  

 

sinA+cosA

Answer 2

Answer:

sinA+cosA

Step-by-step explanation:

cosA/1-(sinA/cosA)+sinA/1-(cosA/sinA)

(where tanA=sinA/cosA,cotA=cosA/sinA)

by lcm we get

cos^2A/cosA-sinA+sin^2A/sinA-cosA

taking -negative common the lcm of denominator becomes cosA-sinA

after lcm we get

cos^2A-sin^2A/cosA-sinA

where cos^2A-sin^2A=(cosA+sinA)(cosA-sinA)

(cosA+sinA)(cosA-sinA)/(cosA-sinA)

cosA-sinA gets cancel from numerator and denominator and we get cosA+sinA

RHS:

sinA+cosA

hence proved LHS=RHS