Answer :

mhanifa

Answer:

  • See below

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Use identities:

  • sin (x + y) = sin x cos y + cos x sin y
  • tan (x - y) =  (tan x - tan y)/(1 + tan x tan y)
  • sin (sin⁻¹ x) = x
  • cos (cos ⁻¹ x) = x
  • tan (tan⁻¹ x) = x
  • sin (cos⁻¹ x) = √ (1 - x²)
  • cos (sin⁻¹ x) = √ (1 - x²)

Question 11

  • sin (sin⁻¹ 4x + cos⁻¹ x) =
  • sin (sin⁻¹ 4x) cos (cos⁻¹ x) + cos (sin⁻¹ 4x) sin(cos⁻¹ x) =
  • 4x*x + √ (1 - (4x)²) * √ (1 - x²)  =
  • 4x² + √(1 - x²)(1 - 16x²)

Question 12

  • tan (tan ⁻¹ 1 - tan ⁻¹ √x) =
  • [tan (tan ⁻¹ 1) - tan (tan ⁻¹ √x)] / [1 + tan (tan ⁻¹ 1) tan (tan ⁻¹ √x)] =
  • (1 - √x) / (1 + 1*√x) =
  • (1 - √x) / (1 + √x)
semsee45

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]