If ${tan}^{2} θ = 1 - a^{2}$ $tan^2theta = 1 - a^2$ , then $sec θ + {tan}^{3} θ csc θ = {(2 - a^{2})}^{n}$ $sectheta + tan^3thetacsctheta = (2 - a^2)^n$ . What is the value of $n$ $n$ ?

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If ${tan}^{2} θ = 1 - a^{2}$ $tan^2theta = 1 - a^2$ , then $sec θ + {tan}^{3} θ csc θ = {(2 - a^{2})}^{n}$ $sectheta + tan^3thetacsctheta = (2 - a^2)^n$ . What is the value of $n$ $n$ ?

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Answer 1 · 2016-11-29T14:28:56

$1.5$ $1.5$

Explanation:

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

Substitute:

$sec θ + {tan}^{3} θ csc θ = {(2 - (1 - {tan}^{2} θ))}^{n}$ $sectheta + tan^3thetacsctheta = (2 - (1 - tan^2theta))^n$

Apply the following identities:

$∙ sec β = \frac{1}{cos β}$ $•secbeta = 1/cosbeta$
$∙ tan β = \frac{sin β}{cos β}$ $•tan beta = sinbeta/cosbeta$
$∙ csc β = \frac{1}{sin β}$ $•cscbeta = 1/sinbeta$

$\frac{1}{cos θ} + \frac{{sin}^{3} θ}{{cos}^{3} θ} \times \frac{1}{sin θ} = {(2 - (1 - {tan}^{2} θ))}^{n}$ $1/costheta + sin^3theta/cos^3theta xx 1/sintheta = (2 - (1 - tan^2theta))^n$

$\frac{1}{cos θ} + \frac{{sin}^{2} θ}{{cos}^{3} θ} = {(1 + {tan}^{2} θ)}^{n}$ $1/costheta + sin^2theta/cos^3theta = (1 + tan^2theta)^n$

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

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

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

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

${sec}^{3} θ = {({sec}^{2} θ)}^{n}$ $sec^3theta = (sec^2theta)^n$

We can call $sec θ = x$ $sectheta = x$ .

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

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

$3 = 2 n$ $3 = 2n$

$n = 1.5$ $n = 1.5$

Hopefully this helps!

Answer 2 · 2016-11-29T14:32:18

$\frac{3}{2}$ $3/2$ and $θ \in Q_{1}$ $theta in Q_1$ .

Explanation:

Let $t = tan θ$ $t = tan theta$ . Then. $t^{2} = 1 - a^{2} \geq 0 \to | a | \leq 1$ $t^2=1-a^2>=0 to |a|<=1$

Now,

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

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

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

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

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

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

$n = \frac{3}{2}$ $n=3/2$ 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 $θ \in Q_{1}$ $theta in Q_1$ .

Answer 3 · 2016-11-29T15:02:18

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

Explanation:

$\frac{1}{cos θ} + \frac{{sin}^{3} θ}{{cos}^{3} θ} \frac{1}{sin θ} = \frac{1}{cos θ} (1 + {tan}^{2} θ) = {(2 - a^{2})}^{n}$ $1/costheta+sin^3theta/cos^3theta 1/sintheta=1/costheta(1+tan^2theta)=(2-a^2)^n$

then

$cos θ = \frac{2 - a^{2}}{{(2 - a^{2})}^{n}} = {(2 - a^{2})}^{- (n - 1)}$ $costheta=(2-a^2)/(2-a^2)^n=(2-a^2)^(-(n-1))$

squaring

${cos}^{2} θ = {(2 - a^{2})}^{- 2 (n - 1)}$ $cos^2theta=(2-a^2)^(-2(n-1))$

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

arriving at

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

$2 (n - 1) = 1 \to n = \frac{3}{2}$ $2(n-1)=1->n=3/2$

Answer 4 · 2016-11-29T15:18:07

Given

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

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

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

Now
$sec θ + {tan}^{3} θ csc θ = {(2 - a^{2})}^{n}$ $sectheta+tan^3thetacsctheta=(2-a^2)^n$

$\Rightarrow sec θ + {tan}^{3} θ \times \frac{1}{sin θ} = {(2 - a^{2})}^{n}$ $=>sectheta+tan^3thetaxx1/sintheta=(2-a^2)^n$

$\Rightarrow sec θ (1 + {tan}^{3} θ \times \frac{cos θ}{sin θ}) = {(2 - a^{2})}^{n}$ $=>sectheta(1+tan^3thetaxxcostheta/sintheta)=(2-a^2)^n$

$\Rightarrow sec θ (1 + {tan}^{3} θ \times \frac{1}{tan θ}) = {({sec}^{2} θ)}^{n}$ $=>sectheta(1+tan^3thetaxx1/tantheta)=(sec^2theta)^n$

$\Rightarrow sec θ (1 + {tan}^{2} θ) = {({sec}^{2} θ)}^{n}$ $=>sectheta(1+tan^2theta)=(sec^2theta)^n$

$\Rightarrow sec θ \times {sec}^{2} θ = {sec}^{2 n} θ$ $=>secthetaxxsec^2theta=sec^(2n)theta$

$\Rightarrow {sec}^{3} θ = {sec}^{2 n} θ$ $=>sec^3theta=sec^(2n)theta$

$\Rightarrow 2 n = 3$ $=>2n=3$

$:.n=3/2$

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If ${tan}^{2} θ = 1 - a^{2}$ $tan^2theta = 1 - a^2$ , then $sec θ + {tan}^{3} θ csc θ = {(2 - a^{2})}^{n}$ $sectheta + tan^3thetacsctheta = (2 - a^2)^n$ . What is the value of $n$ $n$ ?

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If tan^2theta = 1 - a^2tan2θ=1−a2tan^2theta = 1 - a^2, then sectheta + tan^3thetacsctheta = (2 - a^2)^nsecθ+tan3θcscθ=(2−a2)nsectheta + tan^3thetacsctheta = (2 - a^2)^n. What is the value of nnn?

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If ${tan}^{2} θ = 1 - a^{2}$ $tan^2theta = 1 - a^2$ , then $sec θ + {tan}^{3} θ csc θ = {(2 - a^{2})}^{n}$ $sectheta + tan^3thetacsctheta = (2 - a^2)^n$ . What is the value of $n$ $n$ ?