CameraIcon

SearchIcon

MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Modulus of a Complex Number

Express tan...

Question

Express ${tan}^{- 1} (\frac{cos x}{1 - sin x}), - \frac{π}{2} < x < \frac{3 π}{2}$ in the simplest form.

Open in App

Solution

Let us consider the problem:

$t a n^{- 1} (\frac{c o s x}{1 - s i n x}) = t a n^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{\frac{1 - t a n^{2} \frac{x}{2}}{1 + t a n^{2} \frac{x}{2}}}{1 - \frac{2 t a n \frac{x}{2}}{1 + t a n^{2} \frac{x}{2}}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠$

Implies that

$= t a n^{- 1} ⎛ ⎜ ⎜ ⎝ \frac{1 - t a n^{2} \frac{x}{2}}{1 + t a n^{2} \frac{x}{2} - 2 t a n \frac{x}{2}} ⎞ ⎟ ⎟ ⎠$

Implies that

$= t a n^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{(1 + t a n \frac{x}{2}) (1 - t a n \frac{x}{2})}{{(1 - t a n \frac{x}{2})}^{2}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠$

Implies that

$= t a n^{- 1} ⎛ ⎜ ⎜ ⎝ \frac{1 + t a n \frac{x}{2}}{1 - t a n \frac{x}{2}} ⎞ ⎟ ⎟ ⎠$

Implies that

$= t a n^{- 1} ⎛ ⎜ ⎜ ⎝ \frac{t a n \frac{π}{4} + t a n \frac{x}{2}}{1 - t a n \frac{π}{4} t a n \frac{x}{2}} ⎞ ⎟ ⎟ ⎠$

Implies that

$= t a n^{- 1} (t a n (\frac{π}{4} + \frac{x}{2}))$

Hence,

$= \frac{π}{4} + \frac{x}{2}$

Suggest Corrections

thumbs-up

1

Similar questions

{tan}^{- 1} (\frac{cos x}{1 - sin x}), - \frac{π}{2} < x < \frac{π}{2}

in the simplest form.

Q. Write the function

{tan}^{- 1} (\frac{cos x - sin x}{cos x + sin x}), x < π

in the simplest form

{tan}^{- 1} (\frac{\sqrt{1 + cos x} + \sqrt{1 - cos x}}{\sqrt{1 + cos x} - \sqrt{1 - cos x}}) = \frac{π}{4} - \frac{x}{2}

π < x < \frac{3 π}{2}

{tan}^{- 1} (\frac{\sqrt{1 + cos x} + \sqrt{1 - cos x}}{\sqrt{1 + cos x} - \sqrt{1 - cos x}}) = \frac{π}{4} - \frac{x}{2}

π < x < \frac{3 π}{2}

Q. Write the function in the simplest form:

{tan}^{- 1} (\sqrt{\frac{1 - cos x}{1 + cos x}}), x < π

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Geometric Representation and Trigonometric Form

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Modulus of a Complex Number

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon