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

Byju's Answer

Standard XII

Mathematics

General Solutions of (sin theta)^2 = (sin alpha)^2 , (cos theta)^2 = (cos alpha)^2 , (tan theta)^2 = (tan alpha)^2

What is the d...

Question

What is the difference between tan 1 and tan 1 degree?

Open in App

Solution

Suggest Corrections

Similar questions

Sin60 degree = 2tan30 degree/1+tan square30 degree

If $a_{1}, a_{2}, a_{3}, . . ., a_{n}$ is an arithmetic progression with common difference d, then evaluate the following expression.

$t a n [t a n^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + t a n^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + t a n^{- 1} (\frac{d}{1 + a_{3} a_{4}}) + \dots + t a n^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})]$

Q. For the shown circuit, the phase difference between current

I

and

I_{1}

is -

Take

{tan}^{- 1} (\frac{3}{4}) = 37^{\circ}

and

{tan}^{- 1} (\frac{4}{3}) = 53^{\circ}

Q.
lf

a_{1}, a_{2}, a_{3} \dots \dots . . a_{n}

are in A.P. with common difference d, then

t a n [t a n^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + t a n^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + \dots \dots \dots \dots \dots \dots + t a n^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] =

Q. If

a_{1}, a_{2}, a_{3}

,______ are in A.P. of common difference d, then prove that

S = {tan}^{- 1} \frac{d}{1 + a_{1} a_{2}} + {tan}^{- 1} \frac{d}{1 + a_{2} a_{3}} +

_____

+ {tan}^{- 1} \frac{d}{1 + a_{n} a_{n + 1}} = {tan}^{- 1} \frac{n d}{1 + a_{1} a_{n + 1}}

Join BYJU'S Learning Program