Evaluate :
cos 70°sin 20°\dfrac{\text{cos 70°}}{\text{sin 20°}}sin 20°cos 70° + cos 59°sin 31°\dfrac{\text{cos 59°}}{\text{sin 31°}}sin 31°cos 59° - 8 sin2 30°
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cos 70°sin 20°+cos 59°sin 31°−8 sin230°=cos (90° - 70°)sin 20°+cos (90° - 31°)sin 31°−8×(12)2=sin 70°sin 20°+sin 31°sin 31°−8×(14)=sin70°sin20°+sin31°sin31°−2=1+1−2=0\dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 \text{30°}\\[1em] =\dfrac{\text{cos (90° - 70°)}}{\text{sin 20°}} + \dfrac{\text{cos (90° - 31°)}}{\text{sin 31°}} - 8 \times \Big(\dfrac{1}{2}\Big)^2\\[1em] =\dfrac{\text{sin 70°}}{\text{sin 20°}} + \dfrac{\text{sin 31°}}{\text{sin 31°}} - 8 \times \Big(\dfrac{1}{4}\Big)\\[1em] =\dfrac{\cancel{sin 70°}}{\cancel{sin 20°}} + \dfrac{\cancel{sin 31°}}{\cancel{sin 31°}} - 2\\[1em] = 1 + 1 - 2\\[1em] = 0sin 20°cos 70°+sin 31°cos 59°−8 sin230°=sin 20°cos (90° - 70°)+sin 31°cos (90° - 31°)−8×(21)2=sin 20°sin 70°+sin 31°sin 31°−8×(41)=sin20°sin70°+sin31°sin31°−2=1+1−2=0
Hence, cos 70°sin 20°+cos 59°sin 31°−8 sin230°=0\dfrac{\text{cos 70°}}{\text{sin 20°}} + \dfrac{\text{cos 59°}}{\text{sin 31°}} - \text{8 sin}^2 \text{30°} = 0sin 20°cos 70°+sin 31°cos 59°−8 sin230°=0.
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2tan 57°cot 33°2\dfrac{\text{tan 57°}}{\text{cot 33°}}2cot 33°tan 57° - cot 70°tan 20°\dfrac{\text{cot 70°}}{\text{tan 20°}}tan 20°cot 70° - 2cos 45°{\sqrt2} \text{cos 45°}2cos 45°
cot2 41°tan2 49°\dfrac{\text{cot}^2 \text{ 41°}}{\text{tan}^2 \text{ 49°}}tan2 49°cot2 41° - 2 sin2 75° cos2 15°2\dfrac{\text{ sin}^2 \text{ 75°}}{\text{ cos}^2 \text{ 15°}}2 cos2 15° sin2 75°
14 sin 30° + 6 cos 60° - 5 tan 45°.
A triangle ABC is right-angled at B; find the value of sec A . sin C - tan A . tan Csin B\dfrac{\text{sec A . sin C - tan A . tan C}}{\text{sin B}}sin Bsec A . sin C - tan A . tan C.
