Question

Prove that √sec²A+cosec²A = tanA +cotA.

Answer 1

LHS

Root over  sec^2+cosec^2

                   =1/cos^2 + 1/sin^2

                   =(sin^2+cos^2)/(cos^2sin^2)                LCM

                  =root over  1/cos^2sin^2

                  =1/cosAsinA    

 

Rhs

               tanA+cotA

             =sin/cos +  cos/sin

             = (sin^2 +cos^2)/cosAsinA

             =1/cosAsinA

LHS=RHS

Answer 2

Hello users ...
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We know that:
sec²x = 1 + tan²x
And
cosec²x = 1 + cot²x 
And
tan x .cot x = 1 

Solution:-
Taking LHS..
= √(sec²A + cosec²A) 
= √ { (1 + tan²A) + ( 1 + cot²A) } 
= √ ( tan²A + cot²A + 2)
= √  ( tan²A  + cot²A + 2 tan A .cot A ) 

Using 
(a² + b² +2ab) = (a + b)²

= √(tan A + cot A)²
= tan A + cot A = RHS

Hence;
Proved √ (sec²A+cosec²A) = tan A + cot A 
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Hope it helps :)