Question

If sin theta is equal to 2ab by a square plus b square.

Find the value of
(a) cos theta
(b) tan theta
(c) cosec theta
(d) sec theta
(e) cot theta

Answer 1

Solution:

Given that ;

$\sf \sin \theta = \dfrac{2ab}{a {}^{2} + b {}^{2} }$

We know that ;

$\sf \sin \theta = \dfrac{perpendicular}{hypotenuse}$

So, on comparing these two, we get ;

• Perpendicular = 2ab
• Hypotenuse = a² + b²

We need to find base;

$\text{base} = \sqrt{ \text h {}^{2} -\text p {}^{2} } \\ \\ \text{base} = \sqrt{ (\text{a}^{2} + \text b {}^{2} )^{2} -\text{ (2ab})^{2}} \\ \\ \text{base} = \sqrt{ (\text{a}^{2})^{2} + (\text{b} ^{2}) {}^{2} + 2 \text{a} {}^{2} \text{b}^{2} - \text 4 \text{a}^{2} \text{b}^{2} } \\ \\ \text{base} = \sqrt{ (\text{a}^{2})^{2} + (\text{b} ^{2}) {}^{2} - 2 \text{a} {}^{2} \text{b}^{2}} \\ \\ \text{base} = \sqrt{ (\text{a}^{2} - \text b {}^{2} )^{2}} \\ \\ \boxed{ \bf {base } = {a}^{2} - b {}^{2} }$

Now,

$\sf \cos \theta = \frac{b}{h} = \frac{a {}^{2} - {b}^{2} }{ {a}^{2} + {b}^{2} } \\ \\ \sf \tan \theta = \frac{p}{b} = \frac{2ab }{ {a}^{2} - {b}^{2}} \\ \\ \sf \cosec \theta = \frac{h}{p} = \frac{a {}^{2} + {b}^{2} }{ 2ab } \\ \\ \sf \sec \theta = \frac{h}{b} = \frac{a {}^{2} + {b}^{2} }{ {a}^{2} - {b}^{2} } \\ \\ \sf \cot \theta = \frac{b}{p} = \frac{a {}^{2} - {b}^{2} }{ 2ab }$

Answer 2

sin Ø = $\dfrac{2ab}{ {a}^{2} \: + \: {b}^{2} }$

__________ [GIVEN]

We have to find

• cos Ø

• tan Ø

• cosec Ø

• sec Ø

• cot Ø

Here Ø = theta

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sin Ø = $\dfrac{2ab}{ {a}^{2} \: + \: {b}^{2} }$

We know that sin Ø = $\dfrac{P}{H}$

› P = Perpendicular

› H = Hypotenuse

› B = Base

Here..

• P = 2ab

• H = a² + b²

• B = ?

By Pythagoras theorem

=> H² = P² + B²

=> B² = H² - P²

=> B = $\sqrt{ {H }^{2} \: - \: {P}^{2} }$

=> B = $\sqrt{ ({a}^{2} \: + \: {b}^{2} ) \: \: - \: {2ab}^{2} }$

» (a + b)² = a² + 2ab + b²

=> B = $\sqrt{ { ({a}^{2}) }^{2} \: + \: {( {b}^{2} )}^{2} \: + \: 2 {a}^{2} {b}^{2} \: - \: 4 {a}^{2} {b}^{2} }$

=> B = $\sqrt{ { ({a}^{2}) }^{2} \: + \: {( {b}^{2} )}^{2} \: - \: 2 {a}^{2} {b}^{2} }$

» (a - b)² = a² + b² - 2ab

=> B = $\sqrt{ ({a}^{2} \: - \: {b}^{2}) ^{2}}$

=> B = a² - b²

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