formate problem (Question: 470927.0 )

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Kumar Manish

· started a discussion

· 1 Months ago

formate problem (Question: 470927.0 )

question formate problem

Question:

What is the value of $\frac{\tan A - \sin A}{{s i n}^{3} A}$ ?

Options:

$\frac{\sec A}{1 - \cos A}$

$\frac{\sec A}{1 + {c o s}^{2} A}$

$\frac{\sec A}{1 + \cos A}$

None of these

Solution:

Ans: (c)

$\frac{\tan A - \sin A}{{s i n}^{3} A} = \frac{\frac{s i n A}{\cos A} - \sin A}{{s i n}^{3} A}$

= $\frac{s i n A (\frac{1}{\cos A} - 1)}{{s i n}^{3} A} = \frac{1 - \cos A}{\cos A {s i n}^{2} A}$

Rationalising the expression,

= $\frac{(1 - \cos A)}{\cos A . {s i n}^{2} A} \times \frac{(1 + \cos A)}{(1 + c o s A)}$

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

= $\frac{{s i n}^{2} A}{\cos A . {s i n}^{2 A (1 + c o s A)}}$

= $\frac{1}{\cos A} . \frac{1}{1 + c o s A} = \frac{s e c A}{1 + \cos A}$

Alternative:-

$\frac{t a n A - s i n A}{{s i n}^{3} A}$

Put A = 45^O

= $\frac{\tan 45^{0} - \sin 45^{0}}{{s i n}^{3} 45^{0}} = \frac{1 - \frac{1}{\sqrt{2}}}{{(\frac{1}{\sqrt{2}})}^{3}}$

= $\frac{\frac{\sqrt{2} - 1}{\sqrt{2}}}{\frac{1}{2 \sqrt{2}}} = \frac{\sqrt{2 - 1}}{\sqrt{2}} \times 2 \sqrt{2}$

= 2 ( $\sqrt{2}$ - 1)

Now check from options ,

Options:- $\frac{\sec A}{1 + \cos A}$

= $\frac{\sec 45^{0}}{1 + \cos 45^{0}} = \frac{\sqrt{2}}{1 + \frac{1}{\sqrt{2}}}$

= $\frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2} + 1} = \frac{2}{\sqrt{2} + 1}$

Rationalising above quation,

= $\frac{2}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1}$

= $\frac{2 (\sqrt{2} - 1)}{2 - 1} = 2 (\sqrt{2} - 1)$ (Satisfy)

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formate problem (Question: 470927.0 )

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