Question

Please answer this question (trigonometry).

Answer 1

Answer:

L. H. S = (1-sin60)/cos60 = (1-√3/2)/(1/2) = (2-√3)/2/(1/2) = 2-√3

R. H. S = (tan60-1)/(tan60+1) = (√3-1)/(√3+1)

= (√3-1)(√3-1)/(√3+1)(√3-1)

=(√3-1)²/(√3²-1²)

= (√3² -2√3 + 1)/(3-1)

= (3-2√3+1)/2

= (4-2√3)/2

= 2-√3

L. H. S= R. H. S

Answer 2

QuestioN :

Show that ,

$\sf{\frac{1-sin60^\degree}{cos60^\degree}=\frac{tan60^\degree-1}{tan60^\degree+1}}$

SolutioN :

We know,

• sin60° = $\frac{\sqrt{3}}{2}$

• cos60° = $\frac{1}{2}$

• tan60° = $\sqrt{3}$

Taking L.H.S ,

$\sf{\frac{1-sin60^\degree}{cos60^\degree}}\\ \\\sf{=\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}}\\ \\ \sf{=\frac{2-\sqrt{3}}{2}\times\frac{2}{1}}\\ \\ \sf{=2-\sqrt{3}}$

Taking R.H.S,

$\sf{\frac{tan60^\degree-1}{tan60^\degree+1}}\\ \\ \sf{=\frac{\sqrt{3}-1}{\sqrt{3}+1}}\\ \\ \sf{=\frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)(\sqrt{3}-1)}}\\ \\ \sf{=\frac{3-2\sqrt{3}+1}{3-1}}\\ \\ \ sf{=\frac{4-2\sqrt{3}}{2}}\\ \\ \sf{=\frac{2(2-\sqrt{3})}{2}}\\ \\ \sf{=2-\sqrt{3}}$

L.H.S = R.H.S [ Proved ]

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Formulas related to trigonometry :-

• sin²A + cos²A = 1

• 1 + tan²A = sec²A

• 1 + cot²A = cosec²A

• cos²A - sin²A = cos2A

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