Question

Plzzzz help me....... ​

Answer 1

Given :-

An equation, $\bf \dfrac{sin {30}^{ \circ} + tan {45}^{ \circ} - cosec {60}^{ \circ} }{sec {30}^{ \circ} + cos {60}^{ \circ} + cot {45}^{ \circ}}$

What to do ?

To evaluate the given equation, $\bf \dfrac{sin {30}^{ \circ} + tan {45}^{ \circ} - cosec {60}^{ \circ} }{sec {30}^{ \circ} + cos {60}^{ \circ} + cot {45}^{ \circ}}$

Solution :-

Given equation,

$\bf \implies\dfrac{sin {30}^{ \circ} + tan {45}^{ \circ} - cosec {60}^{ \circ} }{sec {30}^{ \circ} + cos {60}^{ \circ} + cot {45}^{ \circ} }$

We know that,

$\bf \bullet \: \: \: sin {30}^{ \circ} = \dfrac{1}{2} \\ \\ \bf \bullet \: \: \: tan {45}^{ \circ} = 1 \\ \\ \bf \bullet \: \: \: cosec {60}^{ \circ} = \dfrac{2}{ \sqrt{3} } \\ \\ \bf \bullet \: \: \: sec {30}^{ \circ} = \dfrac{2}{ \sqrt{3} } \\ \\ \bf \bullet \: \: \: cos {60}^{ \circ} = \dfrac{1}{2} \\ \\ \bf \bullet \: \: \: cot {45}^{ \circ} = 1$

• Substitute all the values in the given equation.

$\bf \implies \dfrac{ \bigg(\dfrac{1}{2} \bigg)+( 1) - \bigg(\dfrac{2}{ \sqrt{3} } \bigg)}{ \bigg( \dfrac{2}{ \sqrt{3} } \bigg) + \bigg( \dfrac{1}{2} \bigg) + 1} \\ \\ \\ \bf \implies \dfrac{ \dfrac{1}{2} + 1 - \dfrac{2}{ \sqrt{3} } }{ \dfrac{2}{ \sqrt{3} }+ \dfrac{1}{2} + 1 } \\ \\ \\ \bf \implies \dfrac{ \dfrac{1 + 2}{2} - \dfrac{2}{ \sqrt{3}}}{ \dfrac{2}{ \sqrt{3}} + \dfrac{1 + 2}{2} }$

$\bf \implies \dfrac{ \dfrac{3}{2} - \dfrac{2}{ \sqrt{3}}}{ \dfrac{2}{ \sqrt{3}} + \dfrac{3}{2} } \\ \\ \\ \bf \implies \dfrac{ \bigg( \dfrac{3 \sqrt{3} - 4}{2 \sqrt{3} } \bigg)}{ \bigg( \dfrac{4 + 3 \sqrt{3}}{2 \sqrt{3} } \bigg)} \\ \\ \\ \bf \implies \dfrac{(3 \sqrt{3}- 4)}{(4 + 3 \sqrt{3})} \times \dfrac{ \not2\sqrt{3} }{ \not2\sqrt{3}}$

$\bf \implies \dfrac{(3 \sqrt{3} - 4)}{(4 + 3 \sqrt{3})} \\ \\ \\ \bf \implies \dfrac{(3 \sqrt{3} - 4)}{(3 \sqrt{3} + 4)} \times \dfrac{(3 \sqrt{3} - 4)}{(3 \sqrt{3} - 4)} \\ \\ \\ \bf \implies \dfrac{ {(3 \sqrt{3} - 4)}^{2} }{(3 \sqrt{3} + 4)(3 \sqrt{3} - 4)}$

• Using two identities in the numerator and the denominator,

(a - b)² = a² - 2ab + b²

(a + b)(a - b) = a² - b²

$\bf \implies \dfrac{ {(3 \sqrt{3})}^{2} - 2(3 \sqrt{3})(4) + {(4)}^{2}}{ {(3 \sqrt{3})}^{2} - {(4)}^{2} } \\ \\ \\ \bf \implies \dfrac{9 \times 3 - 24 \sqrt{3} + 16}{9 \times 3 - 16} \\ \\ \\ \bf \implies \dfrac{27 - 24 \sqrt{3} + 16}{27 - 16} \\ \\ \\ \bf \implies \dfrac{27 + 16 - 24 \sqrt{3}}{11} \\ \\ \\ \bf \implies \dfrac{43 - 24 \sqrt{3} }{11}$

Hence,

$\bf \dfrac{sin {30}^{ \circ} + tan {45}^{ \circ} - cosec {60}^{ \circ} }{sec {30}^{ \circ} + cos {60}^{ \circ} + cot {45}^{ \circ}} = \dfrac{43 - 24 \sqrt{3} }{11}$

Answer 2

Answer:

Solution :-

$\bf \implies \dfrac{ \bigg(\dfrac{1}{2} \bigg)+( 1) - \bigg(\dfrac{2}{ \sqrt{3} } \bigg)}{ \bigg( \dfrac{2}{ \sqrt{3} } \bigg) + \bigg( \dfrac{1}{2} \bigg) + 1} \\ \\ \\ \bf \implies \dfrac{ \dfrac{1}{2} + 1 - \dfrac{2}{ \sqrt{3} } }{ \dfrac{2}{ \sqrt{3} }+ \dfrac{1}{2} + 1 } \\ \\ \\ \bf \implies \dfrac{ \dfrac{1 + 2}{2} - \dfrac{2}{ \sqrt{3}}}{ \dfrac{2}{ \sqrt{3}} + \dfrac{1 + 2}{2} }$

$\bf \implies \dfrac{ \dfrac{3}{2} - \dfrac{2}{ \sqrt{3}}}{ \dfrac{2}{ \sqrt{3}} + \dfrac{3}{2} } \\ \\ \\ \bf \implies \dfrac{ \bigg( \dfrac{3 \sqrt{3} - 4}{2 \sqrt{3} } \bigg)}{ \bigg( \dfrac{4 + 3 \sqrt{3}}{2 \sqrt{3} } \bigg)} \\ \\ \\ \bf \implies \dfrac{(3 \sqrt{3}- 4)}{(4 + 3 \sqrt{3})} \times \dfrac{ \not2\sqrt{3} }{ \not2\sqrt{3}}$

$\bf \implies \dfrac{(3 \sqrt{3} - 4)}{(4 + 3 \sqrt{3})} \\ \\ \\ \bf \implies \dfrac{(3 \sqrt{3} - 4)}{(3 \sqrt{3} + 4)} \times \dfrac{(3 \sqrt{3} - 4)}{(3 \sqrt{3} - 4)} \\ \\ \\ \bf \implies \dfrac{ {(3 \sqrt{3} - 4)}^{2} }{(3 \sqrt{3} + 4)(3 \sqrt{3} - 4)}$

Using two identities in the numerator and the denominator,

(a - b)² = a² - 2ab + b²

(a + b)(a - b) = a² - b²

$\bf \implies \dfrac{ {(3 \sqrt{3})}^{2} - 2(3 \sqrt{3})(4) + {(4)}^{2}}{ {(3 \sqrt{3})}^{2} - {(4)}^{2} } \\ \\ \\ \bf \implies \dfrac{9 \times 3 - 24 \sqrt{3} + 16}{9 \times 3 - 16} \\ \\ \\ \bf \implies \dfrac{27 - 24 \sqrt{3} + 16}{27 - 16} \\ \\ \\ \bf \implies \dfrac{27 + 16 - 24 \sqrt{3}}{11} \\ \\ \\ \bf \implies \dfrac{43 - 24 \sqrt{3} }{11}$

Hence,

$\bf \dfrac{sin {30}^{ \circ} + tan {45}^{ \circ} - cosec {60}^{ \circ} }{sec {30}^{ \circ} + cos {60}^{ \circ} + cot {45}^{ \circ}} = \dfrac{43 - 24 \sqrt{3} }{11}$