Prove that :
cos (2×30°)=1−tan2 30°1+tan2 30°\text{cos }(2 \times 30°) = \dfrac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}cos (2×30°)=1+tan2 30°1−tan2 30°
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L.H.S. = cos (2 x 30°) = cos 60° = 12\dfrac{1}{2}21
R.H.S.
=1−tan2 30°1+tan2 30°=1−(13)21+(13)2=1−131+13=33−1333+13=3−133+13=2343=2343=24=12= \dfrac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}\\[1em] = \dfrac{1 - \Big(\dfrac{1}{\sqrt3}\Big)^2}{1 + \Big(\dfrac{1}{\sqrt3}\Big)^2}\\[1em] = \dfrac{1 - \dfrac{1}{3}}{1 + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{3}{3} - \dfrac{1}{3}}{\dfrac{3}{3} + \dfrac{1}{3}}\\[1em] = \dfrac{\dfrac{3 - 1}{3}}{\dfrac{3 + 1}{3}}\\[1em] = \dfrac{\dfrac{2}{3}}{\dfrac{4}{3}}\\[1em] = \dfrac{\dfrac{2}{\cancel{3}}}{\dfrac{4}{\cancel{3}}}\\[1em] = \dfrac{2}{4}\\[1em] = \dfrac{1}{2}=1+tan2 30°1−tan2 30°=1+(31)21−(31)2=1+311−31=33+3133−31=33+133−1=3432=3432=42=21
∴ L.H.S. = R.H.S.
Hence, cos (2×30°)=1−tan2 30°1+tan2 30°\text{cos }(2 \times 30°) = \dfrac{1 - \text{tan}^2 \text{ 30°}}{1 + \text{tan}^2 \text{ 30°}}cos (2×30°)=1+tan2 30°1−tan2 30°
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3 cosec2 60° - 2 cot2 30° + sec2 45° = 0.
sin (2×30°)=2 tan 30°1+tan2 30°\text{sin }(2 \times 30°) = \dfrac{2 \text{ tan 30°}}{1 + \text{tan}^2 \text{ 30°}}sin (2×30°)=1+tan2 30°2 tan 30°
tan(2×30°)=2 tan 30°1−tan2 30°\text{tan} (2 \times 30°) = \dfrac{\text{2 tan 30°}}{1 - \text{tan}^2 \text{ 30°}}tan(2×30°)=1−tan2 30°2 tan 30°
ABC is an isosceles right-angled triangle. Assuming AB = BC = x, find the value of each of the following trigonometric ratios :
(i) sin 45°
(ii) cos 45°
(iii) tan 45°
