Question

$1 + {tan}^{2} \beta + 1 + \frac{1}{ {tan}^{2} \beta } = \frac{1}{ {sin}^{2} \beta - {sin}^{4 \beta } }$
plz answer this fast......
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Answer 1

Answer:

Note: I am avoiding to write β. Instead I will be answering in terms of A.

L.H.S = 1+tan²A+1+1/tan²A

        = 1 +sin²A/cos²A +1+cos²A/sin²A

        = cos²A+sinA²/cos²A + sin²A+cos²A/sin²A

       =  1/cos²A + 1/sin²A ⇒⇒⇒⇒⇒⇒ (∵ cos²A+sinA=1)

      = sin²A/sin²A*cos²A + cos²A/sin²A*cos²A

       = sin²A+cos²A/sin²A*cos²A

       = 1/sin²A*cos²A ⇒⇒⇒⇒⇒⇒⇒ (∵ cos²A+sinA=1)

       = 1/sin²A*(1-sin²A)⇒⇒⇒⇒⇒⇒ (∵ cos²A = 1-sin²A)

      = 1/sin²A-$sin^{4} A$ = R.H.S

Hope you liked it........

Yours truly,

Shrest Kumar

 

Answer 2

Answer:

Step-by-step explanation:

LHS                                                                        RHS

this is in the form of ( a + b ) ^2                                 1 / sin^2 [ 1 - sin^2 x ]

= (  tan x + 1 /tan x ) ^ 2                                           = 1 /  sin^2x*cos^2x

= tan ^2 x + 1 / tan x  = sec ^4 x  / tan^2x              =    cosec^2 x * sec ^2x

                               =  cosec^2 x * sec ^2x