Secant and Tangent identity

Question

i've been stuck on this question too long

$x = \sec A + \tan A$

show $x + \frac{1}{x} = 2\cdot \sec A$

I've been using $\tan^2 \theta + 1 = \sec^2 \theta$

and $\tan\theta = \frac{\sin \theta}{\cos\theta}$

help would be much appreciated

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$x=\sec A +\tan A = \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A}$

Answer 1

$$ x=\sec A + \tan A= \frac{1}{\cos A}+\frac{\sin A}{\cos A}=\frac{1+\sin A}{\cos A} $$

so we have:

$$ x+\frac{1}{x}= \frac{1+\sin A}{\cos A}+\frac{\cos A}{1+\sin A}=\frac{1+\sin^2 A+2\sin A+\cos^2 A}{(\cos A)(1+ \sin A)}= $$ $$ =\frac{2(1+\sin A)}{(\cos A)(1+ \sin A)}=\frac{2}{\cos A} $$

Answer 2

we get $$x+\frac{1}{x}={\frac { \left( \sec \left( A \right) \right) ^{2}+2\,\sec \left( A \right) \tan \left( A \right) + \left( \tan \left( A \right) \right) ^{2}+1}{\sec \left( A \right) +\tan \left( A \right) }} $$ can you simplify this term? the numerator simplifies to $$-\frac{2}{\sin(A)-1}$$ and $$\sec(A)+\tan(A)=\frac{\sin(A)+1}{\cos(A)}$$

Answer 3

Because $$x+\frac{1}{x}=\frac{1+\sin{A}}{\cos{A}}+\frac{\cos{A}}{1+\sin{A}}=$$ $$=\frac{(1+\sin{A})^2}{\cos{A}(1+\sin{A})}+\frac{\cos^2{A}}{\cos{A}(1+\sin{A})}=\frac{(1+\sin{A})^2+\cos^2A}{\cos{A}(1+\sin{A})}=$$ $$=\frac{1+2\sin{A}+\sin^2A+\cos^2A}{\cos{A}(1+\sin{A})}=\frac{1+2\sin{A}+1}{\cos{A}(1+\sin{A})}=$$ $$=\frac{2+2\sin{A}}{\cos{A}(1+\sin{A})}=\frac{2}{\cos{A}}=2\sec{A}.$$

Answer 4

x=secA + tanA

1/x= 1/(secA + tanA). = (secA - tanA)/(secA-tanA)(secA+tanA) =(secA - tanA)/{(secA)^2 - (tanA)^2} = secA - tanA

Therefore

x+ 1/x = (secA + tanA)+(secA - tanA) = 2SecA

Answer 5

Enough to show:

$$\cos{A}\left( x + \frac{1}{x}\right) = 2$$

$$\cos{A} \cdot x + \frac{\cos{A}}{x} = 1 +\sin{A} + \frac{\cos^2{A}}{1 +\sin{A}} = 1 +\sin{A} + \frac{1-\sin^2{A}}{1 +\sin{A}} = 2 $$