Question

$\sf \red{ Find \: the \: value \: of \: :− }$
secθ(1+sinθ/cosθ+cosθ/1+sinθ)-2tan²θ
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Answer 1

Answer:

Explanation:

Answer : (sinθ  - cosθ + 1) / ( sinθ + cosθ - 1) = 1 /(secθ - tanθ )

Let us deduce the expression from LHS towards RHS

Explanation:

Since we will apply the identity involving secθ and tanθ, first convert the LHS in terms of secθ and tanθ by dividing both numerator and denominator by cosθ.

thus,

LHS = ( sinθ - cosθ +1) / (sinθ + cosθ -1)

       = (sinθ/cosθ +cosθ/cosθ +1/cosθ) / (sinθ/cosθ + cosθ/cosθ -1/cosθ)

       = (tanθ - 1 + secθ) / ( tanθ + 1 - secθ)                              (Since, sinθ/cosθ = tanθ and 1/cosθ = secθ)

       = {( tanθ + secθ) -1} / {(tanθ - secθ) +1}  

     Now multiplying both numerator and denominator by (tanθ - secθ), we get

       = {( tanθ + secθ) -1} (tanθ - secθ) / {(tanθ - secθ) +1}  (tanθ - secθ)

      =  {(tan2θ - sec2θ) - (tanθ - secθ)} / (tanθ - secθ + 1)(tanθ - secθ)

      =  (-1 - tanθ + secθ) / ( tanθ - secθ +1)(tanθ - secθ)

      =      -1 / (tanθ - secθ)

      =      1 / (secθ - tanθ)

      Hence, proved LHS = RHS

Answer 2

Answer:

..Since we will apply the identity involving secθ and tanθ, let us first convert the LHS (of the identity we need to prove) in terms of secθ and tanθ by dividing numerator and denominator by cosθ which is the RHS of the identity, we are required to prove.Read more on Sarthaks.com - https://www.sarthaks.com/256534/prove-that-sin-cos-1-sin-cos-1-1-sec-tan-using-the-identity-sec-2-1-tan-2