Question

if cos 3x=4cos^3x-3cosx, then sec 135 is

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Answer 1

Answer:

-√2

Step-by-step explanation:

Sec135° = ?

= sec(180-45)

= {-sec45}. ....... (since sec is -ve in quadrant 2)

={-1/cos45°}

={-1/(1/√2)}

=[-√2]

Answer 2

The value of sec 135° is -√2.

Explanation:

Given : $\cos 3x=4\cos^3x-3\cos x$    (1)

To find : The value of $\sec{135^{\circ}}$

Consider $\sec{135}=\dfrac{1}{\cos135^{\circ}}\ \ [\because\ \sec A=\dfrac{1}{\cos A}]$

$=\dfrac{1}{ \cos (3(45))^{\circ}}$

$=\dfrac{1}{4\cos^3(45^{\circ}-3\cos (45^{\circ}}$   [From (1)]

$=\dfrac{1}{4(\dfrac{1}{\sqrt{2}})^3-3(\dfrac{1}{\sqrt{2}})}\ \ [\because\ \cos45^{\circ}=\dfrac{1}{\sqrt{2}}]$

$=\dfrac{1}{4(\dfrac{1}{2\sqrt{2}}-\dfrac{3}{\sqrt{2}})}$

$=\dfrac{1}{(\dfrac{2}{\sqrt{2}}-\dfrac{3}{\sqrt{2}})}$

$=\dfrac{1}{\dfrac{2-3}{\sqrt{2}}}$

$=\dfrac{\sqrt{2}}{-1}=-\sqrt{2}$

Hence, the value of $\sec{135^{\circ}}$ is $-\sqrt{2}$.

# Learn more :

Cos²10+Cos²20+Cos²30+Cos²40+Cos²50+Cos²60+Cos²70+Cos²80+Cos²90.

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