Question

If \[\cos \theta = \frac{2}{3}\]  then 2 sec² θ + 2 tan² θ − 7 is equal to 

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Answer 1

Given that:  `cos θ=2/3`

We have to find `2 sec^2 c+2 tan ^2 θ-7`

As we are given 

`cos θ=2/3` 

⇒ `"Base"=2`

⇒ `"Hypotenuse"=3` 

⇒ `"Perpendicular"= sqrt((3)^2-(2)^2)` 

⇒`"Perpendicular"=sqrt5` 

We know that: 

`cos θ="Base"/"Hypotenuse"` 

`tan θ= "Perpendicular"/"Base"` 

Now we have to find:` 2 sec^2θ+2 tan^2 θ-7.` so 

`2 sec^2θ+2 tan ^2 θ-7` 

=`2(3/2)^2+2(sqrt5/2)^2-7` 

= `18/4+10/4-7` 

=`(18+10-28)/4` 

= 0