Answer:
[tex]\textsf{11.} \quad 4x^2+ \sqrt{1-16x^2}\sqrt{1-x^2}[/tex]
[tex]\textsf{12.} \quad \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}[/tex]
Step-by-step explanation:
Question 11
[tex]\boxed{\begin{minipage}{6 cm}\underline{Sine Double Angle Identity}\\\\$\sin (A \pm B)=\sin A \cos B \pm \cos A \sin B$\\\end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{5 cm}\underline{Inverse Trigonometric Identities}\\\\$\sin(\cos^{-1}x)=\sqrt{1-x^2}$\\\\$\cos(\sin^{-1}x)=\sqrt{1-x^2}$\\\end{minipage}}[/tex]
Given trigonometric expression:
[tex]\sin ( \sin^{-1} 4x + \cos^{-1} x)[/tex]
Using the Sine Double Angle Identity:
- [tex]\textsf{Let $A=\sin^{-1}4x$}[/tex]
- [tex]\textsf{Let $B=\cos^{-1}x$}[/tex]
Therefore:
[tex]\begin{aligned}\sin ( \sin^{-1} 4x + \cos^{-1} x)&=\sin(\sin^{-1}4x) \cos (\cos^{-1}x)+ \cos(\sin^{-1}4x) \sin(\cos^{-1}x)\\&=4x \cdot x+ \cos(\sin^{-1}4x) \sin(\cos^{-1}x)\\&=4x^2+ \cos(\sin^{-1}4x) \sin(\cos^{-1}x)\\&=4x^2+ \sqrt{1-(4x)^2}\sqrt{1-x^2}\\&=4x^2+ \sqrt{1-16x^2}\sqrt{1-x^2}\end{aligned}[/tex]
Question 12
[tex]\boxed{\begin{minipage}{5 cm}\underline{Tan Double Angle Identity}\\\\$\tan (A \pm B)=\dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$\\\end{minipage}}[/tex]
Given trigonometric expression:
[tex]\tan(\tan^{-1}1-\tan^{-1}\sqrt{x})[/tex]
Using the Tan Double Angle Identity:
- [tex]\textsf{Let $A=\tan^{-1}1$}[/tex]
- [tex]\textsf{Let $B=\tan^{-1}\sqrt{x}$}[/tex]
Therefore:
[tex]\begin{aligned}\tan(\tan^{-1}1-\tan^{-1}\sqrt{x})&=\dfrac{\tan (\tan^{-1}1) - \tan (\tan^{-1}\sqrt{x})}{1 + \tan (\tan^{-1}1) \tan (\tan^{-1}\sqrt{x})}\\\\&=\dfrac{1 - \sqrt{x}}{1 + 1 \sqrt{x}}\\\\&=\dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}\\\\\end{aligned}[/tex]
