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NCERT Exemplar · Class 10 Mathematics Introduction to Trigonometry and its Applications

60 questions · 60 still being checked

EXERCISE 8.3 1–10 (part 4 of 7)

  1. Prove the following (from Q. $\displaystyle 1$ to Q.7):

    Exercise 1

    sin⁡θ1+cos⁡θ+1+cos⁡θsin⁡θ=2cosec⁡θ\displaystyle \frac{\sin \theta}{1+\cos \theta}+\frac{1+\cos \theta}{\sin \theta}=2 \operatorname{cosec} \theta

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[\frac{\sin\theta}{1+\cos\theta}+\frac{1+\cos\theta}{\sin\theta} = \frac{\sin^2\theta+(1+\cos\theta)^2}{\sin\theta(1+\cos\theta)} \] \[\sin^2\theta+(1+\cos\theta)^2 = \sin^2\theta+\cos^2\theta+2\cos\theta+1 = 2+2\cos\theta \quad (\sin^2\theta+\cos^2\theta=1) \] \[\frac{2+2\cos\theta}{\sin\theta(1+\cos\theta)} = \frac{2(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \frac{2}{\sin\theta} = 2\operatorname{cosec}\theta \]Answer: LHS = RHS = \(\displaystyle 2\operatorname{cosec}\theta\).

  2. Exercise 2

    tan⁡A1+sec⁡A−tan⁡A1−sec⁡A=2cosec⁡A\displaystyle \frac{\tan \mathrm{A}}{1+\sec \mathrm{A}}-\frac{\tan \mathrm{A}}{1-\sec \mathrm{A}}=2 \operatorname{cosec} \mathrm{A}

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[\frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A} = \frac{\tan A(1-\sec A)-\tan A(1+\sec A)}{(1+\sec A)(1-\sec A)} \] \[= \frac{-2\tan A\sec A}{1-\sec^2A} = \frac{-2\tan A\sec A}{-\tan^2A} \quad (1-\sec^2A=-\tan^2A) \] \[= \frac{2\sec A}{\tan A} = \frac{2/\cos A}{\sin A/\cos A} = \frac{2}{\sin A} = 2\operatorname{cosec} A \]Answer: LHS = RHS = \(\displaystyle 2\operatorname{cosec} A\).

  3. Exercise 3

    If tan⁡A=34\displaystyle \tan \mathrm{A}=\frac{3}{4}, then sin⁡Acos⁡A=1225\displaystyle \sin \mathrm{A} \cos \mathrm{A}=\frac{12}{25}

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[\tan A = \frac{3}{4} \] \[\sin A\cos A = \tan A\cos^2A = \frac{\tan A}{\sec^2A} = \frac{\tan A}{1+\tan^2A} \] \[= \frac{3/4}{1+9/16} = \frac{3/4}{25/16} = \frac{3}{4}\times\frac{16}{25} = \frac{12}{25} \]Answer: \(\displaystyle \sin A\cos A = \dfrac{12}{25}\).

  4. Exercise 4

    (sin⁡α+cos⁡α)(tan⁡α+cot⁡α)=sec⁡α+cosec⁡α\displaystyle (\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)=\sec \alpha+\operatorname{cosec} \alpha

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[\tan\alpha+\cot\alpha = \frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha} = \frac{\sin^2\alpha+\cos^2\alpha}{\sin\alpha\cos\alpha} = \frac{1}{\sin\alpha\cos\alpha} \] \[(\sin\alpha+\cos\alpha)(\tan\alpha+\cot\alpha) = \frac{\sin\alpha+\cos\alpha}{\sin\alpha\cos\alpha} = \frac{1}{\cos\alpha}+\frac{1}{\sin\alpha} \] \[= \sec\alpha+\operatorname{cosec}\alpha \]Answer: LHS = RHS = \(\displaystyle \sec\alpha+\operatorname{cosec}\alpha\).

  5. Exercise 5

    (3+1)(3−cot⁡30∘)=tan⁡360∘−2sin⁡60∘\displaystyle (\sqrt{3}+1)\left(3-\cot 30^{\circ}\right)=\tan ^3 60^{\circ}-2 \sin 60^{\circ}

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[\cot 30^\circ=\sqrt3,\quad \tan 60^\circ=\sqrt3,\quad \sin 60^\circ=\frac{\sqrt3}{2} \] \[(\sqrt3+1)(3-\sqrt3) = 3\sqrt3-3+3-\sqrt3 = 2\sqrt3 \] \[\tan^3 60^\circ-2\sin 60^\circ = (\sqrt3)^3-2\cdot\frac{\sqrt3}{2} = 3\sqrt3-\sqrt3 = 2\sqrt3 \]Answer: LHS = RHS = \(\displaystyle 2\sqrt3\).

  6. Exercise 6

    1+cot⁡2α1+cosec⁡α=cosec⁡α\displaystyle 1+\frac{\cot ^2 \alpha}{1+\operatorname{cosec} \alpha}=\operatorname{cosec} \alpha

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[\cot^2\alpha = \operatorname{cosec}^2\alpha-1 = (\operatorname{cosec}\alpha-1)(\operatorname{cosec}\alpha+1) \quad (\operatorname{cosec}^2\alpha-\cot^2\alpha=1) \] \[1+\frac{\cot^2\alpha}{1+\operatorname{cosec}\alpha} = 1+\frac{(\operatorname{cosec}\alpha-1)(\operatorname{cosec}\alpha+1)}{1+\operatorname{cosec}\alpha} = 1+(\operatorname{cosec}\alpha-1) \] \[= \operatorname{cosec}\alpha \]Answer: LHS = RHS = \(\displaystyle \operatorname{cosec}\alpha\).

  7. Exercise 7

    tan⁡θ+tan⁡(90∘−θ)=sec⁡θsec⁡(90∘−θ)\displaystyle \tan \theta+\tan \left(90^{\circ}-\theta\right)=\sec \theta \sec \left(90^{\circ}-\theta\right)

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    \[\tan(90^\circ-\theta)=\cot\theta,\quad \sec(90^\circ-\theta)=\operatorname{cosec}\theta \quad \text{(co-function identities)} \] \[\tan\theta+\cot\theta = \frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} \] \[\sec\theta\operatorname{cosec}\theta = \frac{1}{\cos\theta}\cdot\frac{1}{\sin\theta} = \frac{1}{\sin\theta\cos\theta} \]Answer: LHS = RHS = \(\displaystyle \sec\theta\operatorname{cosec}\theta\).

  8. Exercise 8

    Find the angle of elevation of the sun when the shadow of a pole h\displaystyle h metres high is 3h\displaystyle \sqrt{3} h metres long.

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    NCERT’s answer
    \(\displaystyle 30^{\circ}\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch8_Ex8-3_Q8 \[\tan\theta = \frac{\text{pole}}{\text{shadow}} = \frac{h}{\sqrt3\,h} = \frac{1}{\sqrt3} \] \[\theta = 30^\circ \quad (\tan 30^\circ=\tfrac{1}{\sqrt3}) \]Answer: \(\displaystyle \theta = 30^\circ\).

  9. Exercise 9

    If 3tan⁡θ=1\displaystyle \sqrt{3} \tan \theta=1, then find the value of sin⁡2θ−cos⁡2θ\displaystyle \sin ^2 \theta-\cos ^2 \theta.

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    NCERT’s answer
    \(\displaystyle \frac{-1}{2}\)
    \[\sqrt{3}\tan\theta=1 \quad\Rightarrow\quad \tan\theta=\frac{1}{\sqrt3} \quad\Rightarrow\quad \theta=30^\circ \] \[\sin^2\theta-\cos^2\theta=\left(\frac12\right)^2-\left(\frac{\sqrt3}{2}\right)^2=\frac14-\frac34=-\frac12 \] Answer: \(\displaystyle -\dfrac12\)

  10. Exercise 10

    A ladder 15\displaystyle 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60∘\displaystyle 60^{\circ} with the wall, find the height of the wall.

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    NCERT’s answer
    \(\displaystyle \frac{15}{2} \mathrm{~m}\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch8_Ex8-3_Q10 \[\text{In } \triangle ABC,\ \angle A=60^\circ,\ AC=15\text{ m (ladder)} \] \[\cos 60^\circ=\frac{AB}{AC} \] \[AB=AC\cos 60^\circ=15\times\frac12=7.5\text{ m} \] Answer: \(\displaystyle 7.5\) m