If 2 sin -1-cos -sin -y- then 1-cos -sin -1-sin - is equal to

1 cos α+sin α/1+sin α= 1 cos α+sin α/1+sin α.1+cos α+sin α/1+cos α + sin α=(1+sin α)^2 cos^2α/(1+cos α)(1+cos α+sin`α) =1+2sin α+sin^2α 1+sin^2α/(1+sin α)(1+co ...

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Standard XII

Mathematics

Basic Trigonometric Identities

If 2 sin α/1+...

Question

$I f \frac{2 s i n α}{1 + c o s α + s i n α} = y, t h e n \frac{1 - c o s α + s i n α}{1 + s i n α} i s e q u a l t o$

$\frac{1}{y}$

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$y$

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$1 - y$

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$1 + y$

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Solution

The correct option is B
$y$

$\frac{1 - c o s α + s i n α}{1 + s i n α} = \frac{1 - c o s α + s i n α}{1 + s i n α} . \frac{1 + c o s α + s i n α}{1 + c o s α + s i n α} = \frac{(1 + s i n α)^{2} - c o s^{2} α}{(1 + c o s α) (1 + c o s α + s i n ‘ α)} = \frac{1 + 2 s i n α + s i n^{2} α - 1 + s i n^{2} α}{(1 + s i n α) (1 + c o s α + s i n α)} = \frac{2 s i n α (1 + s i n α)}{(1 + s i n α) (1 + c o s α + s i n α)} = \frac{2 s i n α}{1 + c o s α - s i n α} = y$

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$I f \frac{2 s i n α}{1 + c o s α + s i n α} = y, t h e n \frac{1 - c o s α + s i n α}{1 + s i n α} i s e q u a l t o$

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If 2 sin α1+cos α+sin α=y, then 1−cos α+sin α1+sin α is equal to

The correct option is B y 1−cos α+sin α1+sin α=1−cos α+sinα1+sin α.1+cos α+sin α1+cos α+sin α=(1+sin α)2−cos2α(1+cos α)(1+cos α+sin‘α)=1+2sin α+sin2α−1+sin2α(1+sin α)(1+cos α+sin α)=2 sin α(1+sin α)(1+sin α)(1+cos α+sin α)=2 sin α1+cos α−sin α=y

$I f \frac{2 s i n α}{1 + c o s α + s i n α} = y, t h e n \frac{1 - c o s α + s i n α}{1 + s i n α} i s e q u a l t o$