Question

Evaulate : 4 -
sin 30° +tan 45º - cosec 60°
sec 30° + cos 60° + cos 45º​

Answer 1

Correct Question:

${\sf{Evaluate:\;\dfrac{\sin 30^{\circ}+\tan 45^{\circ}-cosec\;60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}-\cot 45^{\circ}}}}$

Answer:

$\therefore{\boxed{\sf{\dfrac{43-24\sqrt{3}}{11}}}}$

Step-by-step explanation:

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

Now, put the values of following,

$\implies{\sf{\dfrac{\dfrac{1}{2}+1-\dfrac{2}{\sqrt{3}}}{\dfrac{2}{\sqrt{3}} +\dfrac{1}{2} -1}}}$

$\implies{\sf{\dfrac{\dfrac{\sqrt{3}+2\sqrt{3}-4}{2\sqrt{3}}}{\dfrac{4+\sqrt{3}+2\sqrt{3}}{2\sqrt{3}}}}}$

$\implies{\sf{\dfrac{\sqrt{3}+2\sqrt{3}-4}{2\sqrt{3}}\times\dfrac{2\sqrt{3}}{4+\sqrt{3}+2\sqrt{3}}}}$

$\implies{\sf{\dfrac{\sqrt{3}+2\sqrt{3}-4}{4+\sqrt{3}+2\sqrt{3}}}}$

$\implies{\sf{\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}}}$

$\implies{\sf{\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}\times \dfrac{3\sqrt{3}-4}{3\sqrt{3}-4}}}$

$\implies{\sf{\dfrac{(3\sqrt{3}-4)^{2}}{(3\sqrt{3})^{2}-(4)^{2}}}}$

$\implies{\sf{\dfrac{27-12\sqrt{3}-12\sqrt{3}+16}{27-16}}}$

$\implies{\sf{\dfrac{43-24\sqrt{3}}{11}}}$

$\therefore{\boxed{\sf{\dfrac{43-24\sqrt{3}}{11}}}}$

Answer 2

${\large{\underline{\underline{\rm{Question:-}}}}}$

Q) Evaluate :

$\rm \dfrac{sin \: 30 \degree + tan \: 45 \degree - cosec \: 60 \degree}{sec \: 30 \degree + cos \: 60 \degree + cos \: 45 \degree}$

${\large{\underline{\underline{\rm{Solution\checkmark}}}}}$

All we have to do is to put the value of given trigonometric ratios at given angles.

We can find these values by using a trigonometric table.

${\bf{\underline{\underline{Trigonometric\:Table:}}}}$

$\boxed{ \begin{array}{c |c| c |c| c |c} \dfrac{\alpha}{ \quad \quad} & \dfrac{ \bf0 \degree}{ \quad \quad}& \dfrac{ \bf30 \degree}{ \quad \quad}& \dfrac{ \bf45 \degree}{ \quad \quad} & \dfrac{ \bf60 \degree}{ \quad \quad}& \dfrac{ \bf90 \degree}{ \quad \quad} \\ \\ \sf \: sin \: \alpha&0& \pink{\dfrac{1}{2}}& \dfrac{1}{ \sqrt{2}} & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \sf \: cos \: \alpha&1& \dfrac{ \sqrt{3} }{2} & \pink{\dfrac{1}{ \sqrt{2} }} & \pink{\dfrac{1}{2}} &0 \\ \\ \sf \: tan \: \alpha &0& \dfrac{1}{ \sqrt{3} }& \pink1& \sqrt{3} & \rm not \: defined \\ \\ \sf \: cosec \: \alpha& \rm \: not \: defined&2& \sqrt{2} & \pink{\dfrac{2}{ \sqrt{3}} }&1 \\ \\ \sf \: sec \: \alpha &1& \pink{\dfrac{2}{ \sqrt{3} }}& \sqrt{2}&2 & \rm \: not \: defined\end{array}}$

In this table, the required terms are coloured in pink.

Now, putting the values..

$\colon \implies \sf \: \frac{ \frac{1}{2} +1 - \frac{2}{ \sqrt{3} } }{ \frac{2}{ \sqrt{3} } + \frac{ \sqrt{3} }{2} + \frac{1}{ \sqrt{2} } } \\ \\ \colon \implies \: \frac{ \frac{ \sqrt{3 } + 2 \sqrt{3} - 4}{2 \sqrt{3} } }{ \frac{2 \times 2 \sqrt{2} + \sqrt{3} \times \sqrt{3 \times 2} + 2 \sqrt{3} }{2 \sqrt{3 \times 2} } } \\ \\ \colon \implies \sf \: \frac{ \frac{3 \sqrt{3} - 4}{ \cancel{2 \sqrt{3}} } }{ \frac{4 \sqrt{2} + 3 \sqrt{2} + 2 \sqrt{3} }{ \cancel{2 \sqrt{3}} \times \sqrt{2} } } \\ \\ \colon \implies \sf \: \frac{ \sqrt{2}(3 \sqrt{3} - 4) } {7 \sqrt{2} + 2 \sqrt{3} } \\ \\ \colon \implies \sf \: \frac{3 \sqrt{6} - 4 \sqrt{2} }{(7 \sqrt{2} + 2 \sqrt{3} )} \times \frac{(7 \sqrt{2} - 2 \sqrt{3} )}{(7 \sqrt{2} - 2 \sqrt{3} )} \\ \\ \colon \implies \sf \: \frac{7 \sqrt{2}(3 \sqrt{6} - 4 \sqrt{2} ) - 2 \sqrt{3} (3 \sqrt{6} - 4 \sqrt{2} )}{ {(7 \sqrt{2}) }^{2} - {(2 \sqrt{3} )}^{2} } \\ \\ \colon \implies \sf \: \frac{42 \sqrt{3} - 56 - 18 \sqrt{2} + 8 \sqrt{6} }{98 - 12} \\ \\ \colon \implies \sf \: \frac{ \cancel2(21 \sqrt{3} - 28 - 9 \sqrt{2} + 4 \sqrt{6} )}{ \cancel{86}} \\ \\ \colon \implies \boxed{ \red{ \frac{21 \sqrt{3} - 28 - 9 \sqrt{2} + 4 \sqrt{6} }{43} }} \:$

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