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Trigonometric Ratio and Identites question

2017 · Shift 0 · Q32

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Trigonometric Ratio and Identites
2017 · Shift 0 · Q32

Trigonometric Ratio and Identites question

2017 · Shift 0 · Q32

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1

If

5\left( {{{\tan }^2}x - {{\cos }^2}x} \right) = 2\cos 2x + 9

, then the value of

\cos 4x

is :

A
${1 \over 3}$
B
${2 \over 9}$
C
$- {7 \over 9}$
D
$- {3 \over 5}$

View written solutionFree

Correct answer: C

Start with the given equation

$5\left(\tan^2 x-\cos^2 x\right)=2\cos 2x+9$

We want to find $\cos 4x$ .

Express everything in terms of $\cos 2x$

Let

$t=\cos 2x$

Now use the identities:

$\tan^2 x=\frac{1-\cos 2x}{1+\cos 2x}=\frac{1-t}{1+t}$

and

$\cos^2 x=\frac{1+\cos 2x}{2}=\frac{1+t}{2}$

Substitute these into the equation:

$5\left(\frac{1-t}{1+t}-\frac{1+t}{2}\right)=2t+9$

Simplify the left-hand side

Take LCM inside the bracket:

=\frac{2(1-t)-(1+t)^2}{2(1+t)}$$ Now, $$(1+t)^2=1+2t+t^2$$ so numerator becomes $$2-2t-(1+2t+t^2)=1-4t-t^2$$ Thus, $$5\cdot \frac{1-4t-t^2}{2(1+t)}=2t+9$$ 4. Solve for $t$ Multiply both sides by $2(1+t)$: $$5(1-4t-t^2)=2(2t+9)(1+t)$$ Expand both sides: Left side: $$5-20t-5t^2$$ Right side: $$2(2t^2+11t+9)=4t^2+22t+18$$ So, $$5-20t-5t^2=4t^2+22t+18$$ Bring all terms to one side: $$-13-42t-9t^2=0$$ or $$9t^2+42t+13=0$$ Solve: $$t=\frac{-42\pm\sqrt{42^2-4\cdot 9\cdot 13}}{18}$$ $$=\frac{-42\pm\sqrt{1764-468}}{18}$$ $$=\frac{-42\pm\sqrt{1296}}{18}$$ $$=\frac{-42\pm 36}{18}$$ So, $$t=-\frac{1}{3} \quad \text{or} \quad t=-\frac{13}{3}$$ Since $t=\cos 2x$ must lie in $[-1,1]$, we reject $-\frac{13}{3}$. Hence, $$\cos 2x=-\frac{1}{3}$$ 5. Find $\cos 4x$ Using $$\cos 4x=2\cos^2 2x-1$$ we get $$\cos 4x=2\left(\frac{1}{9}\right)-1=\frac{2}{9}-1=-\frac{7}{9}$$ 6. Check options $$\cos 4x=-\frac{7}{9}$$ So the correct option is: **C**.

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