Question

$\sqrt{x} 1+sin\frac{x}{y} 1-sin=sectheeta+tantheeta$

Answer 1

Step-by-step explanation:

Given that we have to prove:

\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}=\frac{\cos \theta}{1-\sin \theta}

tanθ−secθ+1

tanθ+secθ−1

=

1−sinθ

cosθ

Now lets solve left hand side

L . H . S=\frac{\tan \theta+\sec \theta-1}{\tan \theta-\sec \theta+1}L.H.S=

tanθ−secθ+1

tanθ+secθ−1

---- eqn 1

\text { we know } \sec ^{2} \theta-\tan ^{2} \theta=1 we know sec

2

θ−tan

2

θ=1

Apply this in eqn 1 we get

=\frac{\tan \theta+\sec \theta-\left(\sec ^{2} \theta-\tan ^{2} \theta\right)}{\tan \theta-\sec \theta+1}=

tanθ−secθ+1

tanθ+secθ−(sec

2

θ−tan

2

θ)

On expanding we get,

=\frac{\tan \theta+\sec \theta-(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)}{\tan \theta-\sec \theta+1}=

tanθ−secθ+1

tanθ+secθ−(secθ−tanθ)(secθ+tanθ)

=\frac{(\tan \theta+\sec \theta)\{1-(\sec \theta-\tan \theta)\}}{\tan \theta-\sec \theta+1}=

tanθ−secθ+1

(tanθ+secθ){1−(secθ−tanθ)}

=\frac{(\tan \theta+\sec \theta)\{1-\sec \theta+\tan \theta\}}{\tan \theta-\sec \theta+1}=

tanθ−secθ+1

(tanθ+secθ){1−secθ+tanθ}

\begin{lgathered}\begin{array}{l}{=\tan \theta+\sec \theta} \\\\ {=\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}=\frac{\sin \theta+1}{\cos \theta}} \\\\ {\frac{1+\sin \theta}{\cos \theta} \times \frac{1-\sin \theta}{1-\sin \theta}=\frac{1-\sin ^{2} \theta}{\cos \theta(1-\sin \theta)}}\end{array}\end{lgathered}

=tanθ+secθ

=

cosθ

sinθ

+

cosθ

1

=

cosθ

sinθ+1

cosθ

1+sinθ

×

1−sinθ

1−sinθ

=

cosθ(1−sinθ)

1−sin

2

θ

\begin{lgathered}\begin{array}{l}{=\frac{\cos ^{2} \theta}{\cos \theta(1-\sin \theta)}=\frac{\cos \theta}{1-\sin \theta}=\mathrm{R} . \mathrm{H.S}} \\\\ {\text { Hence Proved }}\end{array}\end{lgathered}

=

cosθ(1−sinθ)

cos

2

θ

=

1−sinθ

cosθ

=R.H.S

Hence Proved

Learn more about this topic

Sin theta -cos theta plus one / sin theta plus cos