Prove that tan(theta)tan(60+theta)=tan3theta
The identity is not true,
All I need is one counterexample.
let x=10°
LS = tan10° tan 70° = .48445
RS = tan 30 = .577
LS ≠ RS
we don't have an identity
To prove the given trigonometric identity, we need to manipulate and simplify the expression on the left side of the equation until it matches the expression on the right side.
Starting with the left side of the equation:
tan(theta) * tan(60 + theta)
Let's use the trigonometric identity tan(A + B) = (tan(A) + tan(B)) / (1 - tan(A)tan(B)) to rewrite the expression:
tan(theta) * [tan(60)tan(theta) + tan(theta)tan(theta)] / [1 - tan(60)tan(theta)tan(theta)]
Since tan(60) = √3, we can substitute the values:
tan(theta) * (√3 * tan(theta) + tan(theta)tan(theta)) / [1 - √3 * tan(theta)tan(theta)]
Next, let's use the trigonometric identity tan^2(A) = sec^2(A) - 1 to simplify the expression:
tan(theta) * (√3 * tan(theta) + tan(theta)(sec^2(theta) - 1)) / [1 - √3 * tan(theta)(sec^2(theta) - 1)]
Expanding the expression further:
tan(theta) * (√3 * tan(theta) + tan(theta)sec^2(theta) - tan(theta)) / [1 - √3 * tan(theta)(sec^2(theta) - 1)]
Now we can cancel out some terms:
tan(theta) * (√3 * tan(theta) + tan(theta)sec^2(theta) - tan(theta)) / [1 - √3 * tan(theta)(sec^2(theta) - 1)]
= tan(theta) * (√3 * tan(theta) + tan(theta)(sec^2(theta) - 1)) / [1 - √3 * tan(theta)(sec^2(theta) - 1)]
Further simplifying the expression by factoring out tan(theta):
tan(theta) * [√3 + (sec^2(theta) - 1)] / [1 - √3 * tan(theta)(sec^2(theta) - 1)]
Using the identity sec^2(theta) = 1 + tan^2(theta), we substitute this value in:
tan(theta) * [√3 + (1 + tan^2(theta) - 1)] / [1 - √3 * tan(theta)(1 + tan^2(theta) - 1)]
= tan(theta) * [√3 + tan^2(theta)] / [1 - √3 * tan(theta)]
Finally, we can simplify the expression to match the right side of the equation:
= tan^3(theta) + √3tan(theta)
Therefore, we have successfully proven that tan(theta)tan(60+theta) = tan^3(theta).