Question #2144a

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Question #2144a

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Answer 1 · 2017-12-18T05:22:18

It is proved below:

Explanation:

$tan a + sec a = \frac{x}{y}$ $tan a+sec a=x/y$
$or, \frac{sin a}{cos a} + \frac{1}{cos a} = \frac{x}{y}$ $or,sin a/cos a+1/cos a=x/y$
$or, \frac{sin a + 1}{cos a} = \frac{x}{y}$ $or,(sin a+1)/cos a=x/y$
$or, \frac{{(sin a + 1)}^{2}}{{cos}^{2} a} = \frac{x^{2}}{y^{2}}$ $or,(sin a+1)^2/cos^2 a=x^2/y^2$
$or, \frac{{(sin a + 1)}^{2}}{1 - {sin}^{2} a} = \frac{x^{2}}{y^{2}}$ $or,(sina+1)^2/(1-sin^2a)=x^2/y^2$
$or, \frac{{(sin a + 1)}^{2}}{(1 + sin a) (1 - sin a)} = \frac{x^{2}}{y^{2}}$ $or,(sina+1)^2/((1+sin a)(1-sina))=x^2/y^2$
$or, \frac{sin a + 1}{1 - sin a} = \frac{x^{2}}{y^{2}}$ $or,(sina+1)/(1-sina)=x^2/y^2$
$or, \frac{sin a + 1 - 1 + sin a}{sin a + 1 + 1 - sin a} = \frac{x^{2} - y^{2}}{x^{2} + y^{2}}$ $or,(sina+1-1+sina)/(sina+1+1-sina)=(x^2-y^2)/(x^2+y^2)" "$ [By division-addition method]
$or, \frac{2 sin a}{2} = \frac{x^{2} - y^{2}}{x^{2} + y^{2}}$ $or,(2sina)/(2)=(x^2-y^2)/(x^2+y^2)$
$s o, sin a = \frac{x^{2} - y^{2}}{x^{2} + y^{2}} [P r o v e d .]$ $so,sina=(x^2-y^2)/(x^2+y^2)color(blue)([Proved.])$

Answer 2 · 2017-12-18T07:53:27

Given $sec(a)+tan(a)=x/y......(1)$

We know

$sec^2(a)-tan^2(a)=1$

$=>(sec(a)+tan(a))(sec(a)-tan(a))=1$

$=>sec(a)-sec(a)=1/(sec(a)+tan(a))$

$=>sec(a)-tan(a)=1/(x/y)$

$=>sec(a)-tan(a)=y/x.......(2)$

Adding (1) and (2) we get

$=>2sec(a)=x/y+y/x=(x^2+y^2)/(xy)$

$=>sec(a)=(x^2+y^2)/(2xy)....(3)$

Now subtracting (2) from (1) we get

$=>2tan(a)=x/y-y/x=(x^2-y^2)/(xy)$

$=>tan(a)=(x^2-y^2)/(2xy) .....(4)$

Dividing (4) by (3) we have

$tan(a)/sec(a)=(x^2-y^2)/(x^2+y^2)$

$=>sin(a)=(x^2-y^2)/(x^2+y^2)$

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Question #2144a

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