Trigonometry & Inverse Trigonometry
Trigonometric Ratios and Identities
nta_pyq_2025_apr
Grade 11
Question:
If $\sin x + \sin^2 x = 1$, $x \in \left(0, \dfrac{\pi}{2}\right)$, then $\left(\cos^{12} x + \tan^{12} x\right) + 3\left(\cos^{10} x + \tan^{10} x + \cos^8 x + \tan^8 x\right) + \left(\cos^6 x + \tan^6 x\right)$ is equal to
Step-by-Step Solution
Key Concept: From $\sin x + \sin^2 x = 1$ deduce $\sin x = \cos^2 x$ (so $\tan x = \cos x$); replace all $\tan^{2k} x$ by $\cos^{2k} x$ and factor the resulting polynomial as $2(\sin^2 x + \sin x)^3 = 2(1)^3 = 2$.
$\sin x = 1-\sin^2 x = \cos^2 x$, so $\tan x = \tfrac{\sin x}{\cos x} = \tfrac{\cos^2 x}{\cos x} = \cos x$. The expression becomes $2\cos^{12}x + 6\cos^{10}x + 6\cos^8x + 2\cos^6x = 2\cos^6x(\cos^6x + 3\cos^4x + 3\cos^2x + 1) = 2\cos^6x(\cos^2x+1)^3$. Since $\cos^2x=\sin x$: $\cos^6x=\sin^3x$ and $(\cos^2x+1)^3=(\sin x+1)^3$. Hence the expression $= 2\sin^3x(\sin x+1)^3 = 2[\sin x(\sin x+1)]^3 = 2(\sin^2x+\sin x)^3 = 2(1)^3 = 2$.
Correct Answer: 4