If m-cos A -sin A and n-cos A-sin A- show that m 2 - n 2 m 2 - n 2 - 1 2 A-cos ec A- A -tan A 2-

M=cosA sinA #160; , m=cosA+sinA. m^2+n^2/m^2 n^2 = (cosA sinA)^2(cosA+sinA)^2/(cosA sinA)^2 (cosA+sinA)^2 = cos^2A+sin^2A 2cosA sinA+cos^2A+sin^2A+2c ...

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Byju's Answer

Standard X

Mathematics

Trigonometric Identities

If m=cos A ...

Question

If $m = cos A - sin A$ and $n = cos A + sin A$ ; show that $\frac{m^{2} + n^{2}}{m^{2} - n^{2}} = - \frac{1}{2} sec A . cos e c A = - \frac{(cot A + tan A)}{2}$ .

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Solution

m=cosA-sinA , m=cosA+sinA.
$\frac{m^{2} + n^{2}}{m^{2} - n^{2}} = \frac{(c o s A - s i n A)^{2} (c o s A + s i n A)^{2}}{(c o s A - s i n A)^{2} - (c o s A + s i n A)^{2}}$
$= \frac{c o s^{2} A + s i n^{2} A - 2 c o s A s i n A + c o s^{2} A + s i n^{2} A + 2 c o s A s i n A}{(c o s A - s i n A + c o s A + s i n A) (c o s A - s i n A - c o s A - s i n A)}$
$= \frac{1 + 1}{2 c o s A (- 2 s i n a)}$
$= \frac{2}{- 4 c o s A s i n A}$
$= \frac{- 1}{2} s e c A c o s A .$
$= \frac{- 1}{2 c o s A s i n A}$
$= \frac{- 1}{2 s i n 2 A} [∵ s i n 2 θ = 2 s i n θ c o s θ]$
$= \frac{- 1}{\frac{2 t a n 4}{1 + t a n^{2} A}} [∵ s i n 2 θ = \frac{2 t a n^{2} θ}{1 + t a n^{2} θ}]$
$= \frac{- (1 + t a n^{2} A)}{2 t a n A}$
$= \frac{- 1}{2} (\frac{1}{t a n a} + \frac{t a n^{2} A}{t a n A})$
$= \frac{- 1}{2} (c o t A + t a n A)$

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Similar questions

Q. If

cos A - sin A = m a n d cos A + sin A = n

, show that

\frac{m^{2} - n^{2}}{m^{2} + n^{2}} = - 2 sin A cos A = - \frac{2}{tan A + cot A}

Q. If

tan A + sin A = m

and

tan A - sin A = n

, then
show that

m^{2} - n^{2} = 4 \sqrt{m n}

Q. If

c o s B / s i n A = n

and

c o s B / c o s A = m

Show that

(m^{2} + n^{2}) c o s^{2} A = n^{2}

Q. If

\frac{c o s B}{s i n A} = n a n d \frac{c o s B}{c o s A} = m,

then show that

(m^{2} + n^{2}) c o s^{2} A = n^{2} .

Q. If

m = \frac{sin A}{sin B}, n = \frac{cos A}{cos B}

, then prove that

(m^{2} - n^{2}) {sin}^{2} B = 1 - n^{2}

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If m=cosA−sinA and n=cosA+sinA; show that m2+n2m2−n2=−12secA.cosecA=−(cotA+tanA)2.

If $m = cos A - sin A$ and $n = cos A + sin A$ ; show that $\frac{m^{2} + n^{2}}{m^{2} - n^{2}} = - \frac{1}{2} sec A . cos e c A = - \frac{(cot A + tan A)}{2}$ .