# If tan^2theta = 1 - a^2, then sectheta + tan^3thetacsctheta = (2 - a^2)^n. What is the value of n?

Nov 29, 2016

$1.5$

#### Explanation:

${a}^{2} = 1 - {\tan}^{2} \theta$

Substitute:

$\sec \theta + {\tan}^{3} \theta \csc \theta = {\left(2 - \left(1 - {\tan}^{2} \theta\right)\right)}^{n}$

Apply the following identities:

•secbeta = 1/cosbeta
•tan beta = sinbeta/cosbeta
•cscbeta = 1/sinbeta

$\frac{1}{\cos} \theta + {\sin}^{3} \frac{\theta}{\cos} ^ 3 \theta \times \frac{1}{\sin} \theta = {\left(2 - \left(1 - {\tan}^{2} \theta\right)\right)}^{n}$

$\frac{1}{\cos} \theta + {\sin}^{2} \frac{\theta}{\cos} ^ 3 \theta = {\left(1 + {\tan}^{2} \theta\right)}^{n}$

Put the left hand side on a common denominator and apply the identity $1 + {\tan}^{2} \beta = {\sec}^{2} \beta$.

$\frac{{\cos}^{2} \theta + {\sin}^{2} \theta}{\cos} ^ 3 \theta = {\left({\sec}^{2} \theta\right)}^{n}$

Use the identity ${\sin}^{2} \beta + {\cos}^{2} \beta = 1$.

$\frac{1}{\cos} ^ 3 \theta = {\left({\sec}^{2} \theta\right)}^{n}$

${\sec}^{3} \theta = {\left({\sec}^{2} \theta\right)}^{n}$

We can call $\sec \theta = x$.

${x}^{3} = {\left({x}^{2}\right)}^{n}$

${x}^{3} = {x}^{2 n}$

$3 = 2 n$

$n = 1.5$

Hopefully this helps!

Nov 29, 2016

$\frac{3}{2}$ and $\theta \in {Q}_{1}$.

#### Explanation:

Let $t = \tan \theta$. Then. ${t}^{2} = 1 - {a}^{2} \ge 0 \to | a | \le 1$

Now,

$\sec \theta + {\tan}^{3} \theta \csc \theta$

$= \pm \sqrt{1 + {t}^{2}} + {t}^{3} \left(\pm \sqrt{I + \frac{1}{t} ^ 2}\right)$

$= \pm \sqrt{1 + {t}^{2}} \left(1 \pm {t}^{2}\right)$

$= \pm {\left(1 + {t}^{2}\right)}^{\frac{3}{2}}$, for + sign in the second factor ( $\csc \theta > 0$)

$= \pm {\left(2 - {a}^{2}\right)}^{\frac{3}{2}}$, $-$ sign is taken when $\sec \theta < 0$

$= {\left(2 - {a}^{2}\right)}^{n} \to \sec \theta > 0 \mathmr{and}$

$n = \frac{3}{2}$ and csc theta >0 to theta in Q_1#.

I am sorry that I have involved $\pm$. But, this is Mathematics.

I have obtained an enforcing result $\theta \in {Q}_{1}$.

Nov 29, 2016

$n = \frac{3}{2}$

#### Explanation:

$\frac{1}{\cos} \theta + {\sin}^{3} \frac{\theta}{\cos} ^ 3 \theta \frac{1}{\sin} \theta = \frac{1}{\cos} \theta \left(1 + {\tan}^{2} \theta\right) = {\left(2 - {a}^{2}\right)}^{n}$

then

$\cos \theta = \frac{2 - {a}^{2}}{2 - {a}^{2}} ^ n = {\left(2 - {a}^{2}\right)}^{- \left(n - 1\right)}$

squaring

${\cos}^{2} \theta = {\left(2 - {a}^{2}\right)}^{- 2 \left(n - 1\right)}$

but from ${\tan}^{2} \theta = \left(1 - {a}^{2}\right) \to {\cos}^{2} \theta = {\left(2 - {a}^{2}\right)}^{- 1}$

arriving at

${\left(2 - {a}^{2}\right)}^{- 2 \left(n - 1\right)} = {\left(2 - {a}^{2}\right)}^{- 1}$ so

$2 \left(n - 1\right) = 1 \to n = \frac{3}{2}$

Nov 29, 2016

Given

${\tan}^{2} \theta = 1 - {a}^{2}$

$1 + {\tan}^{2} \theta = 1 + 1 - {a}^{2}$

${\sec}^{2} \theta = 2 - {a}^{2}$

Now
$\sec \theta + {\tan}^{3} \theta \csc \theta = {\left(2 - {a}^{2}\right)}^{n}$

$\implies \sec \theta + {\tan}^{3} \theta \times \frac{1}{\sin} \theta = {\left(2 - {a}^{2}\right)}^{n}$

$\implies \sec \theta \left(1 + {\tan}^{3} \theta \times \cos \frac{\theta}{\sin} \theta\right) = {\left(2 - {a}^{2}\right)}^{n}$

$\implies \sec \theta \left(1 + {\tan}^{3} \theta \times \frac{1}{\tan} \theta\right) = {\left({\sec}^{2} \theta\right)}^{n}$

$\implies \sec \theta \left(1 + {\tan}^{2} \theta\right) = {\left({\sec}^{2} \theta\right)}^{n}$

$\implies \sec \theta \times {\sec}^{2} \theta = {\sec}^{2 n} \theta$

$\implies {\sec}^{3} \theta = {\sec}^{2 n} \theta$

$\implies 2 n = 3$

$\therefore n = \frac{3}{2}$