Find the value of cos A -cos B sin A -sin B n - sin A -sin B cos A -cos B n where n is an even

(cosA+cosB/sinA sinB)^n+(sinA+sinB/cosA cosB)^n #160; = (2cos(A+B/2)cos(A B/2)/2cos(A B/2)sin(A B/2))^n+(2sin(A+B/2)cos(A B/2)/ 2sin(A+B/2)sin(A B/2)) ...

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Byju's Answer

Standard XII

Mathematics

Basic Trigonometric Identities

Find the valu...

Question

Find the value of ${(\frac{cos A + cos B}{sin A - sin B})}^{n} + {(\frac{sin A + sin B}{cos A - cos B})}^{n}$ (where $n$ is an even)

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Solution

$(\frac{c o s A + c o s B}{s i n A - s i n B})^{n} + (\frac{s i n A + s i n B}{c o s A - c o s B})^{n}$
$= (\frac{2 c o s (\frac{A + B}{2}) c o s (\frac{A - B}{2})}{2 c o s (\frac{A - B}{2}) s i n (\frac{A - B}{2})})^{n} + (\frac{2 s i n (\frac{A + B}{2}) c o s (\frac{A - B}{2})}{- 2 s i n (\frac{A + B}{2}) s i n (\frac{A - B}{2})})^{n}$
$= c o t^{n} (\frac{A - B}{2}) + (- 1)^{n} c o t^{n} (\frac{A - B}{2})$
if n is even, than $(- 1)^{n} = 1$ then the given expression becomes
$2 c o t^{n} (\frac{A - B}{2})$
If n is odd then $(- 1)^{n} = - 1$ then the given expression becomes 0

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Similar questions

Q. The value of

{(\frac{cos A + cos B}{sin A - sin B})}^{n} + {(\frac{sin A + sin B}{cos A - cos B})}^{n}

Q. If

n

is an odd integer then

{(\frac{cos A + cos B}{sin A - sin B})}^{n} + {(\frac{sin A + sin B}{cos A - cos B})}^{n} =

Q. Assertion :

{(\frac{cos A + cos B}{sin A - sin B})}^{n} + {(\frac{sin A + sin B}{cos A - cos B})}^{n}

= 2 {cot}^{n} (\frac{A - B}{2})

n

is even

= 0

n

is odd. Reason:

\frac{cos A + cos B}{sin A - sin B} = cot \frac{A - B}{2}

Q. lf n is an odd integer, then

{(\frac{cos A + cos B}{sin A - sin B})}^{n} + {(\frac{sin A + sin B}{cos A - cos B})}^{n}

is equal to

{(\frac{cos A + cos B}{sin A - sin B})}^{n} + {(\frac{sin A + sin B}{cos A - cos B})}^{n} = x {cot}^{n} \frac{A - B}{2}

or a, accordingly as n is even or odd. Find

x + a

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Find the value of (cosA+cosBsinA−sinB)n+(sinA+sinBcosA−cosB)n (where n is an even)

Find the value of ${(\frac{cos A + cos B}{sin A - sin B})}^{n} + {(\frac{sin A + sin B}{cos A - cos B})}^{n}$ (where $n$ is an even)