\(\frac{11sin70° }{7cos20° } - \frac{4cos53°cosec37°}{7tan15° tan35° tan75°tan55° }\)
According to the trigonometric ratio of complementry angles:
If θ is acute:
sin(90 - θ) = cosθ, cos(90 - θ) = sinθ, tan(90 - θ) = cotθ,
sec(90 - θ) = cosecθ, cosec(90 - θ) = secθ, cot(90 - θ) = tanθ,
means, sin(70) = sin(90 - 20) = cos20
cos(53) = cos(90 - 37) = sin(37)
tan15 = tan(90 - 75) = cot (75)
and, tan(35) = cot(55),
Now on putting in given expression
\(\frac{11sin(20) }{7cos(20) } - \frac{4cos(37)cosec(37)}{7tan(75) tan(55) tan(75)tan(55)}\)
as, tanθ x cotθ = 1, we get
\(\frac{11} 7 - \frac47 = 1\)
