Question

plz solve the (x) part..

Answer 1

Hey there !!

Taking LHS

(1+tan²A)
------------- 
(1+cot²A)

We know that cot²A=1/tan²A


(1+tan²A)      (1+tan²A)
------------- =   -------------
(1+cot²A)       (1+1/tan²A)
               
                =(1+tan²A)
                  -------------
                 (tan²A+1)
                 -------------
                  tan²A
              
=  tan²A

Therefore Prooved that LHS =RHS

Now 

Similarly

 (1-tanA)²   
------------   
(1-cotA)²

writing tanA as 1/cotA and simplifying we get (1-tanA)²/(1-cotA)²=tan²A

Answer 2

[(1+tan²A)/(1+cot²A)]

(As we know →

sec²A = 1 + tan²A

cosec²A= 1+ cot²A )

So, when we put the values →

sec²A / cosec²A

(As we know →

sec²A = 1/cos²A

cosec²A = 1/sin²A)

When we put the values →

(1/cos²A)/(1/sin²A)

= sin²A/cos²A

= tan²A

Hence , LHS = RHS

So, proved.