Evaluate:
tan 47°cot 43°\dfrac{\text{tan 47°}}{\text{cot 43°}}cot 43°tan 47°
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tan 47°cot 43°=tan (90° - 43°)cot 43°=cot 43°cot 43°=cot43°cot43°=1\dfrac{\text{tan 47°}}{\text{cot 43°}} = \dfrac{\text{tan (90° - 43°)}}{\text{cot 43°}}\\[1em] = \dfrac{\text{cot 43°}}{\text{cot 43°}}\\[1em] = \dfrac{\cancel{cot 43°}}{\cancel{cot 43°}}\\[1em] = 1cot 43°tan 47°=cot 43°tan (90° - 43°)=cot 43°cot 43°=cot43°cot43°=1
Hence, tan 47°cot 43°\dfrac{\text{tan 47°}}{\text{cot 43°}}cot 43°tan 47° = 1.
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In a triangle ABC, secA+C2\text{sec}\dfrac{A + C}{2}sec2A+C is equal to:
0
secB2\text{sec}\dfrac{B}{2}sec2B
cosec B
cosecB2\text{cosec}\dfrac{B}{2}cosec2B
cos 22°sin 68°\dfrac{\text{cos 22°}}{\text{sin 68°}}sin 68°cos 22°
sec 75°cosec 15°\dfrac{\text{sec 75°}}{\text{cosec 15°}}cosec 15°sec 75°
cos 55°sin 35°\dfrac{\text{cos 55°}}{\text{sin 35°}}sin 35°cos 55° + cot 35°tan 55°\dfrac{\text{cot 35°}}{\text{tan 55°}}tan 55°cot 35°
