Question

List-I	List-II
(I) The value of tan 20°. tan 40°. tan 60°. tan 80° is	(P) Odd natural
(II) The value of 4$(sin^4 \frac{\pi}{16} + sin^4 \frac{3\pi}{16} + sin^4 \frac{5\pi}{16} + sin^4 \frac{7\pi}{16})$ is	(Q) Even natural
(III) tana + 2tan2a + 4tan4a + 8cot8a = $\lambda$cota then $\lambda$ is	(R) 3
(IV) (4 cos²9° -3) (4 cos²27° -3) = tank°, then k is	(S) 6
	(T) 1
	(U) 9

Which of the following is CORRECT?

Answer 1

I-R, II-S, III-T, IV-U

Answer 2

The problem requires us to evaluate four trigonometric expressions and match their results to the given options.

(I) The value of tan 20°. tan 40°. tan 60°. tan 80° is We use the trigonometric identity: $\tan \theta \tan(60^\circ - \theta) \tan(60^\circ + \theta) = \tan(3\theta)$. Let $\theta = 20^\circ$. Then, $\tan 20^\circ \tan(60^\circ - 20^\circ) \tan(60^\circ + 20^\circ) = \tan 20^\circ \tan 40^\circ \tan 80^\circ$. According to the identity, this product is equal to $\tan(3 \times 20^\circ) = \tan 60^\circ = \sqrt{3}$. The full expression is $\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = (\tan 20^\circ \tan 40^\circ \tan 80^\circ) \times \tan 60^\circ$. Substituting the values: $\sqrt{3} \times \sqrt{3} = 3$. So, (I) matches with (R).

(II) The value of 4$(sin^4 \frac{\pi}{16} + sin^4 \frac{3\pi}{16} + sin^4 \frac{5\pi}{16} + sin^4 \frac{7\pi}{16})$ is Let's analyze the angles: $\frac{7\pi}{16} = \frac{8\pi - \pi}{16} = \frac{\pi}{2} - \frac{\pi}{16}$. So, $\sin(\frac{7\pi}{16}) = \cos(\frac{\pi}{16})$. $\frac{5\pi}{16} = \frac{8\pi - 3\pi}{16} = \frac{\pi}{2} - \frac{3\pi}{16}$. So, $\sin(\frac{5\pi}{16}) = \cos(\frac{3\pi}{16})$. Substitute these into the expression: $4(\sin^4 \frac{\pi}{16} + \sin^4 \frac{3\pi}{16} + \cos^4 \frac{3\pi}{16} + \cos^4 \frac{\pi}{16})$ Rearrange the terms: $4[(\sin^4 \frac{\pi}{16} + \cos^4 \frac{\pi}{16}) + (\sin^4 \frac{3\pi}{16} + \cos^4 \frac{3\pi}{16})]$ We use the identity $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}(2\sin x \cos x)^2 = 1 - \frac{1}{2}\sin^2(2x)$. For $x = \frac{\pi}{16}$: $\sin^4 \frac{\pi}{16} + \cos^4 \frac{\pi}{16} = 1 - \frac{1}{2}\sin^2(\frac{2\pi}{16}) = 1 - \frac{1}{2}\sin^2(\frac{\pi}{8})$. For $x = \frac{3\pi}{16}$: $\sin^4 \frac{3\pi}{16} + \cos^4 \frac{3\pi}{16} = 1 - \frac{1}{2}\sin^2(\frac{6\pi}{16}) = 1 - \frac{1}{2}\sin^2(\frac{3\pi}{8})$. Substitute these back: $4[(1 - \frac{1}{2}\sin^2(\frac{\pi}{8})) + (1 - \frac{1}{2}\sin^2(\frac{3\pi}{8}))]$ $4[2 - \frac{1}{2}(\sin^2(\frac{\pi}{8}) + \sin^2(\frac{3\pi}{8}))]$ $8 - 2(\sin^2(\frac{\pi}{8}) + \sin^2(\frac{3\pi}{8}))$ Now, notice that $\frac{3\pi}{8} = \frac{4\pi - \pi}{8} = \frac{\pi}{2} - \frac{\pi}{8}$. So, $\sin(\frac{3\pi}{8}) = \cos(\frac{\pi}{8})$. Therefore, $\sin^2(\frac{3\pi}{8}) = \cos^2(\frac{\pi}{8})$. The expression becomes: $8 - 2(\sin^2(\frac{\pi}{8}) + \cos^2(\frac{\pi}{8}))$ $8 - 2(1)$ (since $\sin^2 x + \cos^2 x = 1$) $8 - 2 = 6$. So, (II) matches with (S).

(III) tana + 2tan2a + 4tan4a + 8cot8a = $\lambda$cota then $\lambda$ is We use the identity: $\cot x - \tan x = 2\cot 2x$. This can be rearranged as $\tan x = \cot x - 2\cot 2x$. Let's simplify the given expression from right to left: Consider the term $8\cot 8a$. From the identity $2\cot 2x = \cot x - \tan x$, let $x=4a$. Then $2\cot 8a = \cot 4a - \tan 4a$. So, $8\cot 8a = 4(2\cot 8a) = 4(\cot 4a - \tan 4a)$. Substitute this into the expression: $tana + 2tan2a + 4tan4a + 4(\cot 4a - \tan 4a)$ $= tana + 2tan2a + 4tan4a + 4\cot 4a - 4\tan 4a$ $= tana + 2tan2a + 4\cot 4a$ Now consider $4\cot 4a = 2(2\cot 4a)$. Let $x=2a$. $2\cot 4a = \cot 2a - \tan 2a$. So, $4\cot 4a = 2(\cot 2a - \tan 2a)$. Substitute this back: $tana + 2tan2a + 2(\cot 2a - \tan 2a)$ $= tana + 2tan2a + 2\cot 2a - 2\tan 2a$ $= tana + 2\cot 2a$ Finally, consider $2\cot 2a$. Let $x=a$. $2\cot 2a = \cot a - \tan a$. Substitute this back: $tana + (\cot a - \tan a)$ $= \cot a$. Comparing with $\lambda \cot a$, we find $\lambda = 1$. So, (III) matches with (T).

(IV) (4 cos²9° -3) (4 cos²27° -3) = tank°, then k is We use the triple angle formula for cosine: $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$. Dividing by $\cos\theta$ (assuming $\cos\theta \neq 0$): $\frac{\cos 3\theta}{\cos \theta} = 4\cos^2 \theta - 3$. For the first term, let $\theta = 9^\circ$: $4\cos^2 9^\circ - 3 = \frac{\cos(3 \times 9^\circ)}{\cos 9^\circ} = \frac{\cos 27^\circ}{\cos 9^\circ}$. For the second term, let $\theta = 27^\circ$: $4\cos^2 27^\circ - 3 = \frac{\cos(3 \times 27^\circ)}{\cos 27^\circ} = \frac{\cos 81^\circ}{\cos 27^\circ}$. Multiply these two terms: $(\frac{\cos 27^\circ}{\cos 9^\circ}) \times (\frac{\cos 81^\circ}{\cos 27^\circ})$ The $\cos 27^\circ$ terms cancel out, leaving: $\frac{\cos 81^\circ}{\cos 9^\circ}$. We know that $\cos 81^\circ = \cos(90^\circ - 9^\circ) = \sin 9^\circ$. So the expression becomes $\frac{\sin 9^\circ}{\cos 9^\circ} = \tan 9^\circ$. Comparing this with $\tan k^\circ$, we get $k = 9$. So, (IV) matches with (U).

Final Matches: (I) - (R) (II) - (S) (III) - (T) (IV) - (U)

The question asks "Which of the following is CORRECT?". Since no options are provided in the prompt, we assume the correct answer is the set of these pairings.