Question 1, Review Exercise 10

Solutions of Question 1 of Review Exercise 10 of Unit 10: Trigonometric Identities of Sum and Difference of Angles. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Chose the correct option.

i. $\cos {{50}^{\circ }}5{0}'\cos {{9}^{\circ }}1{0}'-\sin {{50}^{\circ }}5{0}'\sin {{9}^{\circ }}1{0}'=$

• (a) $0$
• (b) $\dfrac{1}{2}$
• (c) $1$
• (d) $\dfrac{\sqrt{3}}{2}$
  See Answer

  (B): $\dfrac{1}{2}$

ii. If $\tan {{15}^{\circ }}=2-\sqrt{3}$, then the value of ${{\cot }^{2}}{{75}^{\circ }}$ is

• (a) $7+\sqrt{3}$
• (b) $7-2\sqrt{3}$
• (c) $7-4\sqrt{3}$
• (d) $7+4\sqrt{3}$
  See Answer

  (B): $\dfrac{1}{2}$

iii. If $\tan \left( \alpha +\beta \right)=\dfrac{1}{2}$, and $\tan \alpha =\dfrac{1}{3}$ then $\tan \beta =$

• (a) $\dfrac{1}{6}$
• (b) $\dfrac{1}{7}$
• (c) $1$
• (d) $\dfrac{7}{6}$
  See Answer

  (B): $\dfrac{1}{2}$

iv. $\sin \theta \cos \left( {{90}^{\circ }}-\theta \right)+\cos \theta \sin \left( {{90}^{\circ }}-\theta \right)=$

• (a) $-1$
• (b) $2$
• (c) $0$
• (d) $1$
  See Answer

  (B): $\dfrac{1}{2}$

v. Simplified expression of $\left( \sec \theta +\tan \theta \right)\left( 1-\sin \theta \right)$ is

• (a) ${{\sin }^{2}}\theta$
• (b) ${{\cos }^{2}}\theta$
• (c) $ta{{n}^{2}}\theta$
• (d) $\cos \theta$
  See Answer

  (B): $\dfrac{1}{2}$

vi. $\sin \left( x-\frac{\pi }{2} \right)=$ is

• (a) $\sin x$
• (b) $-\sin x$
• (c) $\cos x$
• (d) $-\cos x$
  See Answer

  (B): $\dfrac{1}{2}$

vii. A point is in Quadrant-III and on the unit circle. If its x-coordinate is $-\dfrac{4}{5},$ what is the y-coordinate of the point?

• (a) $\dfrac{3}{5}$
• (b) $-\dfrac{3}{5}$
• (c) $-\dfrac{2}{5}$
• (d) $\dfrac{5}{3}$
  See Answer

  (B): $\dfrac{1}{2}$

viii. Which of the following is an identity?

• (a) $\sin \left( a \right)\cos \left( a \right)=\left( \dfrac{1}{2} \right)\left( \sin 2a \right)$
• (b) $\sin a+\cos a=1$
• (c) $\sin \left( -a \right)=\sin a$
• (d) $\tan a=\dfrac{\cos a}{\sin a}$
  See Answer

  (a): $\sin \left( a \right)\cos \left( a \right)=\left( \dfrac{1}{2} \right)\left( \sin 2a \right)$

Question 2 & 3 >