Question

prove that tan theta + sec theta -1 / tan theta -sec theta +1 =cos theta/ 1 - sin theta

plz reply fast if possible

Answer 1

Question :-

Prove that -

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

Solution :-

Here, L.H.S. is  $\frac{tan \theta + sec \theta - 1}{tan \theta - sec \theta + 1}$

and R.H.S. is  $\frac{cos\theta}{1-sin\theta}$

• Solving the L.H.S.,

$=>\frac{tan \theta + sec \theta - 1}{tan \theta - sec \theta + 1}$

Replacing 1 by sec²Ф - tan²Ф.

As, [sec²Ф - tan²Ф = 1]

$=>\frac{tan \theta + sec \theta -(sec^{2} \theta - tan^{2} \theta )}{tan \theta - sec \theta + 1}\\\\=>\frac{(sec \theta + tan \theta) - (sec \theta + tan \theta)(sec \theta - tan \theta)}{tan \theta - sec \theta + 1}\\\\=>\frac{(sec \theta + tan \theta)(1-(sec \theta - tan \theta))}{tan \theta - sec \theta + 1}\\\\=>\frac{(sec \theta + tan \theta)(1-sec \theta + tan \theta)}{tan \theta - sec \theta + 1}$

Cancelling the like terms i.e., (1-secФ+tanФ)

$=> sec\theta+tan\theta$

We know that, [ $sec\theta= \frac{1}{cos\theta} \:and\: tan\theta = \frac{sin \theta}{cos \theta}$ ]

So,

=> $\frac{1}{cos\theta} + \frac{sin \theta}{cos \theta}$

=> $\frac{1-sin\theta}{cos\theta}$  = R.H.S.

L.H.S. = R.H.S.

Hence proved.

More Information :-

$\star$ cot θ = cos θ / sin θ

$\star$ cot θ = 1 / tan θ

$\star$ tan θ = sin θ / cos θ

$\star$ tan θ = 1 / cos θ

$\star$ sin² θ + cos² θ =1

$\star$ sec² θ - tan² θ =1

Answer 2

Question :-

Prove that -

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

Solution :-

Here, L.H.S. is  $\frac{tan \theta + sec \theta - 1}{tan \theta - sec \theta + 1}$

and R.H.S. is  $\frac{cos\theta}{1-sin\theta}$

• Solving the L.H.S.,

$=>\frac{tan \theta + sec \theta - 1}{tan \theta - sec \theta + 1}$

Replacing 1 by sec²Ф - tan²Ф.

As, [sec²Ф - tan²Ф = 1]

$=>\frac{tan \theta + sec \theta -(sec^{2} \theta - tan^{2} \theta )}{tan \theta - sec \theta + 1}\\\\=>\frac{(sec \theta + tan \theta) - (sec \theta + tan \theta)(sec \theta - tan \theta)}{tan \theta - sec \theta + 1}\\\\=>\frac{(sec \theta + tan \theta)(1-(sec \theta - tan \theta))}{tan \theta - sec \theta + 1}\\\\=>\frac{(sec \theta + tan \theta)(1-sec \theta + tan \theta)}{tan \theta - sec \theta + 1}$

Cancelling the like terms i.e., (1-secФ+tanФ)

$=> sec\theta+tan\theta$

We know that, [ $sec\theta= \frac{1}{cos\theta} \:and\: tan\theta = \frac{sin \theta}{cos \theta}$ ]

So,

=> $\frac{1}{cos\theta} + \frac{sin \theta}{cos \theta}$

=> $\frac{1-sin\theta}{cos\theta}$  = R.H.S.

L.H.S. = R.H.S.

Hence proved.