Question

f 1-tan2°cot62°/tan152°-cot88°=k√3 then k=

Answer 1

Given :-

$\sf \: \frac{1 - tan2 ° cot62° \: }{tan152° - cot88° \: } = k \sqrt{3}$

To find :-

• the value of k

solution :-

$:\implies\sf \: \frac{1 - tan2 ° cot62° \: }{tan152° - cot88° \: } = k \sqrt{3} \\ \\ \\ :\implies\sf \: \frac{1 - \: tan2° \: cot(90° - 28°)}{tan(180° \: - 28°) - cot(90° \: - 2°} = k \sqrt{3} \\ \\ \\ :\implies\sf \: \frac{1 - tan2°tan28° \: }{ - tan28° - tan2° \: } = k \sqrt{3}$

$\implies\sf \: \frac{ \frac{ - 1}{tan28 + tan2} }{1 - tan2 \: tan28} \\ \\ \\ \implies\sf \: \frac{ - 1}{tan(28 + 2)} = k \sqrt{3} \\ \\ \\ \implies\sf \: \frac{ - 1}{tan30} = k \sqrt{3} \\ \\ \\ \implies\sf \: - cot30 = k \sqrt{3}$

$\implies\sf \: - \sqrt{3} = k \sqrt{3} \\ \\ \\ \implies\sf \therefore \: k = - 1$

so therefore value of k is -1