Question

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Answer 1

Answer:

$\dfrac{3 \sqrt{2} - \sqrt{6} }{8}$

Step-by-step explanation:

$\red{\boxed{\rm Solution}}$

$\dfrac{ \cos(45) }{ \sec(30) + \csc(30) }$

Here csc means cosec

We know that secθ = 1/cosθ and cosecθ = 1/sinθ

We also know that cos(45) = 1/√2

Therefore,

$\dfrac{ \dfrac{1}{ \sqrt{2} } }{ \dfrac{1}{ \cos(30) } + \dfrac{1}{ \sin(30) } }$

We know that cos 30 = √3/2 and sin 30 = 1/2 from the table

$\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3} }{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \rm Not \: De fined \\ \\ \rm cosec A & \rm Not \: De fined & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm Not \: De fined \\ \\ \rm cot A & \rm Not \: De fined & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}$

Simplifying it more,

$\dfrac{ \dfrac{1}{ \sqrt{2} } }{ \dfrac{1}{\dfrac{ \sqrt{3} }{2}} + \dfrac{1}{ \dfrac{1}{2} } }$

$\dfrac{ \dfrac{1}{ \sqrt{2} } }{ \dfrac{2}{ \sqrt{3} } + 2 }$

$\dfrac{ \dfrac{1}{ \sqrt{2} } }{ \dfrac{2 + 2 \sqrt{3} }{ \sqrt{3} } }$

$\dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2 + 2 \sqrt{3} }$

Rationalise the denominators

$\dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{2} }{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2(1 + \sqrt{3}) } \times \dfrac{(1 - \sqrt{3}) }{(1 - \sqrt{3}) }$

$\dfrac{ \sqrt{2} }{ { \sqrt{2} }^{2} } \times \dfrac{ \sqrt{3}(1 - \sqrt{3}) }{2(1 + \sqrt{3})(1 - \sqrt{3} ) }$

We know that (√n)² = n, as square and square root gets cancelled

We also know that (a + b)(a - b) = a² - b²

Applying the above formulas

$\dfrac{ \sqrt{2} }{2} \times \dfrac{ \sqrt{3} - 3 }{2(1 - 3)}$

$\dfrac{ \sqrt{2} }{2} \times \dfrac{ \sqrt{3} - 3 }{ - 4}$

$\dfrac{ \sqrt{2}( \sqrt{3} - 3) }{ - 8}$

$\dfrac{ \sqrt{2}(3 - \sqrt{3}) }{8}$

We know that √a x √b = √ab

Applying the above formula,

$\dfrac{3 \sqrt{2} - \sqrt{6} }{8}$