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Home  >>  CBSE XI  >>  Math  >>  Trigonometric Functions

If $A$ and $B$ are positive acute angle and $\cos A=\large\frac{1}{7},$$\cos B=\large\frac{13}{14}$,then show that $A-B=60^{\large\circ}$

$\begin{array}{1 1}(A)\;\large\frac{1}{2}&(B)\;\large\frac{1}{4}\\(C)\;\large\frac{1}{6}&(D)\;\large\frac{1}{8}\end{array} $

1 Answer

Toolbox:
  • $\cos(A-B)=\cos A\cos B+\sin A.\sin B$
  • $\sin A=\sqrt{1-\cos^2A}$
  • $\sin B=\sqrt{1-\cos^2B}$
Since $\cos A=\large\frac{1}{7}$ & $\cos B=\large\frac{13}{14}$
$\therefore \sin A=\sqrt{1-\cos A}$
$\Rightarrow \sqrt{1-(\large\frac{1}{7})^2}=\large\frac{4\sqrt 3}{7}$
$\sin B=\sqrt{1-\cos^2B}=\sqrt{1-(\large\frac{13}{15})^2}=\large\frac{3\sqrt 3}{14}$
$\therefore\cos(A-B)=\cos A\cos B+\sin A.\sin B$
$\Rightarrow \large\frac{1}{7}.\frac{13}{14}+\frac{4\sqrt 3}{7}.\frac{3\sqrt 3}{14}$
$\Rightarrow \large\frac{49}{7\times 14}=\frac{1}{2}$
Hence (A) is the correct answer.
answered May 21, 2014 by sreemathi.v
 

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