Question

Evaluate sin 30 degree by cos 45 degree + cot 45 degree by sec 60 degree minus sin 60 degree by tan 45 degree minus cos 60 degree by sin 90 degree

Answer 1

$\textbf{To find:}$

$\text{The value of}$

$\dfrac{sin\,30^{\circ}}{cos\,45^{\circ}}+\dfrac{cot\,45^{\circ}}{sec\,60^{\circ}}-\dfrac{sin\,60^{\circ}}{tan\,45^{\circ}}-\dfrac{cos\,60^{\circ}}{sin\,90^{\circ}}$

$\textbf{Solution:}$

$\textbf{Standard trigonometric table:}$

$\left\begin{array}{|c|c|c|c|c|c|}\cline{1-6}&0^{\circ}&30^{\circ}&45^{\circ}&60^{\circ}&90^{\circ}\\\cline{1-6}\bf\,sin\theta&0&\frac{1}{2}&\frac{1}{\sqrt2}&\frac{\sqrt3}{2}&1\\\cline{1-6}\bf\,cos\theta&1&\frac{\sqrt3}{2}&\frac{1}{\sqrt2}&\frac{1}{2}&0\\\cline{1-6}\bf\,tan\theta&0&\frac{1}{\sqrt3}&1&\sqrt3&\infty\\\cline{1-6}\end{array}\right$

$\dfrac{sin\,30^{\circ}}{cos\,45^{\circ}}+\dfrac{cot\,45^{\circ}}{sec\,60^{\circ}}-\dfrac{sin\,60^{\circ}}{tan\,45^{\circ}}-\dfrac{cos\,60^{\circ}}{sin\,90^{\circ}}$

$=\dfrac{\frac{1}{2}}{\frac{1}{\sqrt2}}+\dfrac{1}{sec\,60^{\circ}}-\dfrac{\frac{\sqrt3}{2}}{1}-\dfrac{\frac{1}{2}}{1}$

$=\dfrac{\sqrt2}{2}+cos\,60^{\circ}-\dfrac{\sqrt3}{2}-\dfrac{1}{2}$

$=\dfrac{1}{\sqrt2}+\dfrac{1}{2}-\dfrac{\sqrt3}{2}-\dfrac{1}{2}$

$=\dfrac{1}{\sqrt2}-\dfrac{\sqrt3}{2}$

$=\dfrac{2-\sqrt{6}}{2\sqrt2}$

$\therefore\textbf{The value of \bf\,\dfrac{sin\,30^{\circ}}{cos\,45^{\circ}}+\dfrac{cot\,45^{\circ}}{sec\,60^{\circ}}-\dfrac{sin\,60^{\circ}}{tan\,45^{\circ}}-\dfrac{cos\,60^{\circ}}{sin\,90^{\circ}} is \bf\dfrac{2-\sqrt{6}}{2\sqrt2} }$

Find more:

1.Solve without using trigonometric tables

(tan 20 / cot 70)+(cot 50 / tan 40 ) +((sin2 20 + sin2 70 )/ sin theta cos (90 - theta ) +cos theta sin (90 - theta)

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2.Cos(90-theta)sec(90-theta)tan(theta)upon cosec(90-theta)sin(90-theta)cot(90-theta)+tan(90-theta)upon cot(theta) =2

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Answer 2

Answer:

sin 30°/cos 45° + cot 45°/sec 60° - sin 60°/ tan 45°+ cos 30° / sin 90°

(1/2) / (1/√2) + 1/2 - (√3/2) / 1 + (√3/2) / 1

√2/2 + 1/2 -√3/2 + √3/2

√2 + 1/2 Hence proved