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Prove the Identity: tan^2 x + 6 tanx + 5 / sec^2 x - 2 = tanx + 5 / tanx - 1?
2 Answers
- 4 years ago
Hint: sec(x)^2 - tan(x)^2 = 1
tan(x)^2 + 6 * tan(x) + 5 =>
(tan(x) + 5) * (tan(x) + 1)
sec(x)^2 - 2 =>
1 + tan(x)^2 - 2 =>
tan(x)^2 - 1 =>
(tan(x) - 1) * (tan(x) + 1)
You should be able to finish this.
- az_lenderLv 74 years ago
Maybe you mean
[tan^2(x) + 6 tan(x) + 5] / [sec^2(x) - 2]
= [tan(x) + 5] / [tan(x) - 1].
When you replace a horizontal fraction bar with a diagonal slash, symbols of inclusion (parentheses or brackets) are absolutely REQUIRED.
Anyway, since
tan^2(x) + 6 tan(x) + 5
= [tan(x) + 5] * [tan(x) + 1],
your equality reduces to
[tan(x) + 1] / [sec^2(x) - 2] = 1/[tan(x) - 1], or
[tan^2(x) - 1] = [sec^2(x) - 2],
but this is obvious because
tan^2(x) = sec^2(x) - 1.
