If cos B -sin A - n and cos B -cos A - m- then show that m 2- n 2 cos 2 A - n 2-

N = cos B/sin A ; m = cos B/cos A So, n^2 = cos^2 B/sin^2 A; m^2 = cos^2 B/cos^2 A L.H.S. = (m^2 + n^2)cos^2 A = (cos^2 B/cos^2 A + cos^2 B/sin^2 A) cos^2 A = ...

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Standard X

Mathematics

Visualisation of Trigonometric Ratios Using a Unit Circle

If cos B /sin...

Question

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

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Solution

$n = \frac{c o s B}{s i n A}; m = \frac{c o s B}{c o s A}$
So, $n^{2} = \frac{c o s^{2} B}{s i n^{2} A}; m^{2} = \frac{c o s^{2} B}{c o s^{2} A}$
$L . H . S . = (m^{2} + n^{2}) c o s^{2} A = (\frac{c o s^{2} B}{c o s^{2} A} + \frac{c o s^{2} B}{s i n^{2} A}) c o s^{2} A = \frac{s i n^{2} A c o s^{2} B + c o s^{2} A c o s^{2} B}{c o s^{2} A s i n^{2} A} \times c o s^{2} A = \frac{c o s^{2} B (s i n^{2} A + c o s^{2} A)}{s i n^{2} A} = \frac{c o s^{2} B}{s i n^{2} A}$
$= n^{2} = R . H . S .$ Hence Proved.

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If cos Bsin A=n and cos Bcos A=m, then show that (m2+n2)cos2A=n2.

n=cos Bsin A;m=cos Bcos A So, n2=cos2Bsin2A; m2=cos2Bcos2A L.H.S.=(m2+n2)cos2A=(cos2Bcos2A+cos2Bsin2A)cos2A=sin2A cos2B+cos2A cos2Bcos2A sin2A×cos2A=cos2B(sin2A+cos2A)sin2A=cos2Bsin2A =n2=R.H.S. Hence Proved.

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