Question

sin theta+cos theta/sin theta - costheta + sin theta-cos theta/sin theta+cos theta =2sec^2 theta/tan^2-1=2/2sin^2theta-1​

Answer 1

Answer:

Solution:

\begin{gathered}\frac{sin \: \theta + cos \: \theta}{sin \: \theta - cos \: \theta} + \frac{sin \: \theta - cos \: \theta}{sin \: \theta + cos \: \theta} \\ \\ \frac{ {(sin \: \theta + cos \: \theta)}^{2} +{(sin \: \theta - cos \: \theta)}^{2} }{ { {sin}^{2} \theta - {cos}^{2}\theta } } \\ \\ \frac{{sin}^{2} \theta + {cos}^{2} \theta + 2sin \: \theta \: cos \: \theta +{sin}^{2} \theta + {cos}^{2} \theta - 2sin \: \theta \: cos \: \theta }{{sin}^{2} \theta - {cos}^{2}\theta} \\ \\\end{gathered}

sinθ−cosθ

sinθ+cosθ

+

sinθ+cosθ

sinθ−cosθ

sin

2

θ−cos

2

θ

(sinθ+cosθ)

2

+(sinθ−cosθ)

2

sin

2

θ−cos

2

θ

sin

2

θ+cos

2

θ+2sinθcosθ+sin

2

θ+cos

2

θ−2sinθcosθ

Since

\begin{gathered}{sin}^{2} \theta + {cos}^{2} \theta = 1 \\ \\\end{gathered}

sin

2

θ+cos

2

θ=1

So

\begin{gathered}\frac{1 + 1}{ {sin}^{2}\theta - {cos}^{2}\theta } \\ \\ = \frac{2}{ {cos}^{2}\theta\bigg( \frac{ {sin}^{2}\theta }{ {cos}^{2}\theta } - 1\bigg) } \\ \\\end{gathered}

sin

2

θ−cos

2

θ

1+1

=

cos

2

θ(

cos

2

θ

sin

2

θ

−1)

2

As

\begin{gathered}\frac{1}{cos \: \theta} = sec \: \theta \\ \\ \frac{sin \: \theta}{cos \:\theta} \: = tan \: \theta \\ \\\end{gathered}

cosθ

1

=secθ

cosθ

sinθ

=tanθ

So

\begin{gathered}= \frac{2 {sec}^{2}\theta }{ {tan}^{2}\theta - 1 } \\ \\ hence \: proved \\ \\\end{gathered}

=

tan

2

θ−1

2sec

2

θ

hence proved

Hope it helps you.

Step-by-step explanation:

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