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`(sin 70^@ + cos 40^@)/(cos 70^@ + sin 40^@)` |
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Answer» `(sin70^@ +cos40^@)/(cos70^@+sin40^@)` We know, `sintheta = cos(90-theta)`. So, our expression becomes `(cos20^@+cos40^@)/(sin20^@+sin40^@)` As ,`cosA+cosB = 2cos((A+B)/2)cos((A-B)/2)` and `sinA+sinB = 2sin((A+B)/2)cos((A-B)/2)` Our expression becomes, `=(2cos((20^@+40^@)/2)cos((20^@-40^@)/2))/(2sin((20^@+40^@)/2)cos((20^@-40^@)/2))` `=cot30^@ = sqrt3` |
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