Chapter 8 · Introduction to Trigonometry and its Applications

NCERT Exemplar Class 10 Maths Exercise 8.3

All 15 questions of NCERT Exemplar Class 10 Maths Exercise 8.3 (Introduction to Trigonometry and its Applications), in the book's order, each with its answer and a step-by-step solution.

1. Q1
   Prove ​that $\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=2\mathrm{cosec}\theta$.
   Show answer

   LHS simplifies to $\frac{2}{\sin \theta }=2\mathrm{cosec}\theta$, proving the identity.

   Step-by-step solution →
2. Q2
   Prove ​that $\frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}=2\mathrm{cosec}A$.
   Show answer

   LHS simplifies to $2\mathrm{cosec}A$, which equals the RHS. Hence the identity is proved.

   Step-by-step solution →
3. Q3
   If $\tan A=\frac{3}{4}$, ​then $\sin A\cos A=\frac{12}{25}$.
   Show answer

   True

   Step-by-step solution →
4. Q4
   Prove ​that $(\sin \alpha +\cos \alpha )(\tan \alpha +\cot \alpha )=\sec \alpha +\csc \alpha$.
   Show answer

   LHS $=\sec \alpha +\csc \alpha =$ RHS (proved).

   Step-by-step solution →
5. Q5
   Prove ​that $\left(\sqrt{3}+1\right)(3-\cot 30^{\circ })={\tan }^{3}60^{\circ }-2\sin 60^{\circ }$.
   Show answer

   Both sides equal $2\sqrt{3}$, so the identity is proved.

   Step-by-step solution →
6. Q6
   Prove ​that $1+\frac{{\cot }^2\alpha }{1+\csc \alpha }=\csc \alpha$.
   Show answer

   The left side simplifies to $\csc \alpha$, equal to the right side, so the identity is proved.

   Step-by-step solution →
7. Q7
   Prove ​that $\tan \theta +\tan (90^{\circ }-\theta )=\sec \theta \sec (90^{\circ }-\theta )$.
   Show answer

   Proved: $\tan \theta +\tan (90^{\circ }-\theta )=\frac{1}{\sin \theta \cos \theta }=\sec \theta \sec (90^{\circ }-\theta )$.

   Step-by-step solution →
8. Q8
   Find the ​angle of elevation of the ​sun ​when the ​shadow of ​a pole $h$ metres high is $\sqrt{3}h$ metres long.
   Show answer

   The angle of elevation of the sun is $30^{\circ }$.

   Step-by-step solution →
9. Q9
   If $\sqrt{3}\tan \theta =1$, ​then find the ​value ​of ${\sin }^2\theta -{\cos }^2\theta$.
   Show answer

   ${\sin }^2\theta -{\cos }^2\theta =-\frac{1}{2}$ (with $\theta =30^{\circ }$).

   Step-by-step solution →
10. Q10
    A ladder 15 metres ​long ​just ​reaches the top ​of ​a vertical wall. If the ​ladder makes an angle of $60^{\circ }$ with ​the wall, find the height of the wall.
    Show answer

    The height of the wall is $15\times \cos 60^{\circ }=7.5$ m.

    Step-by-step solution →
11. Q11
    Simplify $(1+{\tan }^2\theta )(1-\sin \theta )(1+\sin \theta )$.
    Show answer

    $(1+{\tan }^2\theta )(1-\sin \theta )(1+\sin \theta )=1$.

    Step-by-step solution →
12. Q12
    If $2{\sin }^2\theta -{\cos }^2\theta =2$, ​then ​find ​the ​value ​of $\theta$.
    Show answer

    $\theta =90^{\circ }$.

    Step-by-step solution →
13. Q13
    Show ​that $\frac{{\cos }^2(45^{\circ }+\theta )+{\cos }^2(45^{\circ }-\theta )}{\tan (60^{\circ }+\theta )\tan (30^{\circ }-\theta )}=1$.
    Show answer

    The expression equals $1$ (proved).

    Step-by-step solution →
14. Q14
    An observer $1.5$ metres tall is $20.5$ metres away from a ​tower $22$ ​metres high. Determine the angle of elevation ​of ​the ​top of ​the tower from the eye of the ​observer.
    Show answer

    The angle of elevation is $45^{\circ }$.

    Step-by-step solution →
15. Q15
    Show ​that ${\tan }^4\theta +{\tan }^2\theta ={\sec }^4\theta -{\sec }^2\theta$.
    Show answer

    Proved: ${\sec }^4\theta -{\sec }^2\theta ={\sec }^2\theta ({\sec }^2\theta -1)=(1+{\tan }^2\theta ){\tan }^2\theta ={\tan }^4\theta +{\tan }^2\theta$.

    Step-by-step solution →