Question

Evaluate the the following (tan20/cosec70)^2 +(cot 20/sec70)^2 +2tan15 tan45 tan 75. Please help

Answer 1

Given That
(tan20/cosec70)^2+(cot20/sec70)^2+2tan15 tan45 tan75 =?
We Know That
(1/cosec70) =sin70 =sin(90-20) =cos20
(1/sec70) =cos70 =cos(90-20) =sin20
(tan75) =tan(90-15) =cot15
These Above Are Finded From The Properties Of Triangles
Then Comes To The Given Problem
Let X Be The Required Value Then
X=(tan20/cosec70)^2+(cot20/sec70)^2+2 tan15 tan45 tan75
Substitute The Finded Values In The Above
Then
X=(tan20 cos20) ^2+(cot20 sin20) ^2+2 tan15 tan45 cot15
We Know That tan20=sin20/cos20
cot20=cos20/sin20
Then
X=(sin20)^2+(cos20) ^2+2 tan15 tan45 cot15
X=1+2=3
Because Of (sinx)^2+(cosx) ^2=1
And tanx cotx =1
Then The Required Value Is 3