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12 (a) Show that the equation $$2 \cot^2 x + 2 \csc^2 x = 1 + 4 \csc x$$ can be written in the form a \csc^2 x + b \csc x + c = 0 12 (b) Hence, given $x$ is obtuse and $$2 \cot^2 x + 2 \csc^2 x = 1 + 4 \csc x$$ find the exact value of $\tan x$ - AQA - A-Level Maths Pure - Question 12 - 2019 - Paper 1

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Question 12

12-(a)-Show-that-the-equation--$$2-\cot^2-x-+-2-\csc^2-x-=-1-+-4-\csc-x$$--can-be-written-in-the-form--a-\csc^2-x-+-b-\csc-x-+-c-=-0---12-(b)-Hence,-given-$x$-is-obtuse-and--$$2-\cot^2-x-+-2-\csc^2-x-=-1-+-4-\csc-x$$--find-the-exact-value-of-$\tan-x$-AQA-A-Level Maths Pure-Question 12-2019-Paper 1.png

12 (a) Show that the equation $$2 \cot^2 x + 2 \csc^2 x = 1 + 4 \csc x$$ can be written in the form a \csc^2 x + b \csc x + c = 0 12 (b) Hence, given $x$ is obt... show full transcript

Worked Solution & Example Answer:12 (a) Show that the equation $$2 \cot^2 x + 2 \csc^2 x = 1 + 4 \csc x$$ can be written in the form a \csc^2 x + b \csc x + c = 0 12 (b) Hence, given $x$ is obtuse and $$2 \cot^2 x + 2 \csc^2 x = 1 + 4 \csc x$$ find the exact value of $\tan x$ - AQA - A-Level Maths Pure - Question 12 - 2019 - Paper 1

Step 1

Show that the equation $2 \cot^2 x + 2 \csc^2 x = 1 + 4 \csc x$ can be written in the form $a \csc^2 x + b \csc x + c = 0$

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Answer

To rewrite the equation, we start by using the Pythagorean identity for cotangent:

cot2x=csc2x1\cot^2 x = \csc^2 x - 1

Substituting this into the equation:

2(csc2x1)+2csc2x=1+4cscx2(\csc^2 x - 1) + 2 \csc^2 x = 1 + 4 \csc x

This simplifies to:

2csc2x2+2csc2x=1+4cscx2\csc^2 x - 2 + 2\csc^2 x = 1 + 4 \csc x

Combining like terms yields:

4csc2x4=1+4cscx4\csc^2 x - 4 = 1 + 4\csc x

Rearranging gives:

4csc2x4cscx5=04\csc^2 x - 4\csc x - 5 = 0

Thus, we can identify:

  • a=4a = 4
  • b=4b = -4
  • c=5c = -5.

Step 2

Find the exact value of $\tan x$

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Answer

Given that xx is obtuse, we can analyze the quadratic equation obtained in part (a):

2cot2x+2csc2x=1+4cscx2 \cot^2 x + 2 \csc^2 x = 1 + 4 \csc x

From our previous work, we found:

cscx=32orcscx=1\csc x = \frac{3}{2} \, \text{or} \, \csc x = -1

The valid solution must satisfy cscx1|\csc x| \geq 1, therefore, we take:

cscx=32\csc x = \frac{3}{2}

The definition of cosecant is given by:

cscx=1sinx\csc x = \frac{1}{\sin x}

Thus:

sinx=23\sin x = \frac{2}{3}

Now, using the identity:

tanx=sinxcosx\tan x = \frac{\sin x}{\cos x}

And recalling that for an obtuse angle, we have:

cosx=1sin2x\cos x = -\sqrt{1 - \sin^2 x}

Calculating this gives:

cosx=1(23)2=149=59=53\cos x = -\sqrt{1 - \left(\frac{2}{3}\right)^2} = -\sqrt{1 - \frac{4}{9}} = -\sqrt{\frac{5}{9}} = -\frac{\sqrt{5}}{3}

Now substituting back into the tangent formula:

tanx=2353=25=255\tan x = \frac{\frac{2}{3}}{-\frac{\sqrt{5}}{3}} = -\frac{2}{\sqrt{5}} = -\frac{2\sqrt{5}}{5}

Thus, the exact value of tanx\tan x is:

tanx=255\tan x = -\frac{2\sqrt{5}}{5}

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