If cos- x-sin- cot- sin x-cos- - then the value of tan- x-2 is

The given equation can be written as 1 tan^2 ( x/2 )/1+tan^2 ( x/2 ) sinα cotβ2tan ( x/2 )/1+tan^2 ( x/2 )=cos α ⇒ tan^2 x/2(1+cosα )+sinα cotβ .2tanx/2 (1 cos ...

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Byju's Answer

Standard XIII

Mathematics

Trigonometric Equations

If cos⁡ x-sin...

Question

If $c o s x - s i n α c o t β s i n x = c o s α$ ,
then the value of $t a n (\frac{x}{2})$ is

- t a n (\frac{α}{2}) c o t (\frac{β}{2})

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t a n (\frac{α}{2}) t a n (\frac{β}{2})

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- c o t (\frac{α}{2}) t a n (\frac{β}{2})

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c o t (\frac{α}{2}) t a n (\frac{β}{2})

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Solution

The correct option is B $t a n (\frac{α}{2}) t a n (\frac{β}{2})$
The given equation can be written as
$\frac{1 - t a n^{2} (\frac{x}{2})}{1 + t a n^{2} (\frac{x}{2})} - s i n α c o t β \frac{2 t a n (\frac{x}{2})}{1 + t a n^{2} (\frac{x}{2})} = c o s α \Rightarrow t a n^{2} \frac{x}{2} (1 + c o s α) + s i n α c o t β .2 t a n \frac{x}{2} - (1 - c o s α) = 0 \Rightarrow t a n^{2} \frac{x}{2} + \frac{2 s i n α c o t β}{1 + c o s α} t a n \frac{x}{2} - \frac{1 - c o s α}{1 + c o s α} = 0 \Rightarrow t a n^{2} \frac{x}{2} + 2 t a n \frac{α}{2} c o t β t a n \frac{x}{2} - t a n^{2} \frac{α}{2} = 0 \Rightarrow t a n^{2} \frac{x}{2} + 2 t a n \frac{α}{2} . \frac{1}{2} (c o t \frac{β}{2} - t a n \frac{β}{2}) t a n \frac{x}{2} - t a n^{2} \frac{α}{2} = 0 \Rightarrow (t a n \frac{x}{2} + c o t \frac{β}{2} t a n \frac{α}{2}) (t a n \frac{x}{2} - t a n \frac{β}{2} t a n \frac{α}{2}) = 0 \Rightarrow t a n (\frac{x}{2}) = - t a n (\frac{α}{2}) c o t (\frac{β}{2}) o r t a n (\frac{x}{2}) = t a n (\frac{α}{2}) t a n (\frac{β}{2})$

Suggest Corrections

If cos x−sinαcotβsin x=cosα , then the value of tan(x2) is

If $c o s x - s i n α c o t β s i n x = c o s α$ ,
then the value of $t a n (\frac{x}{2})$ is