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

60 questions · 60 still being checked

EXERCISE 8.3 11–15 (part 5 of 7)

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

    Exercise 11

    Simplify (1+tan⁡2θ)(1−sin⁡θ)(1+sin⁡θ)\displaystyle \left(1+\tan ^2 \theta\right)(1-\sin \theta)(1+\sin \theta)

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    NCERT’s answer
    $\displaystyle 1$
    \[(1+\tan^2\theta)(1-\sin\theta)(1+\sin\theta)=\sec^2\theta\,(1-\sin^2\theta) \] \[=\sec^2\theta\cos^2\theta=1 \] Answer: \(\displaystyle 1\)

  2. Exercise 12

    If 2sin⁡2θ−cos⁡2θ=2\displaystyle 2 \sin ^2 \theta-\cos ^2 \theta=2, then find the value of θ\displaystyle \theta.

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    NCERT’s answer
    \(\displaystyle 90^{\circ}\)
    \[2\sin^2\theta-\cos^2\theta=2 \] \[2\sin^2\theta-(1-\sin^2\theta)=2 \] \[3\sin^2\theta=3 \quad\Rightarrow\quad \sin^2\theta=1 \quad\Rightarrow\quad \sin\theta=1 \] \[\theta=90^\circ \] Answer: \(\displaystyle \theta=90^\circ\)

  3. Exercise 13

    Show that cos⁡2(45∘+θ)+cos⁡2(45∘−θ)tan⁡(60∘+θ)tan⁡(30∘−θ)=1\displaystyle \frac{\cos ^2\left(45^{\circ}+\theta\right)+\cos ^2\left(45^{\circ}-\theta\right)}{\tan \left(60^{\circ}+\theta\right) \tan \left(30^{\circ}-\theta\right)}=1

    Not cross-checked

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

    \[\cos(45^\circ-\theta)=\cos\big(90^\circ-(45^\circ+\theta)\big)=\sin(45^\circ+\theta) \] \[\cos^2(45^\circ+\theta)+\cos^2(45^\circ-\theta)=\cos^2(45^\circ+\theta)+\sin^2(45^\circ+\theta)=1 \] \[\tan(30^\circ-\theta)=\tan\big(90^\circ-(60^\circ+\theta)\big)=\cot(60^\circ+\theta) \] \[\tan(60^\circ+\theta)\tan(30^\circ-\theta)=\tan(60^\circ+\theta)\cot(60^\circ+\theta)=1 \] \[\frac{\cos^2(45^\circ+\theta)+\cos^2(45^\circ-\theta)}{\tan(60^\circ+\theta)\tan(30^\circ-\theta)}=\frac{1}{1}=1 \] Answer: Proved.

  4. Exercise 14

    An observer 1.5\displaystyle 1.5 metres tall is 20.5\displaystyle 20.5 metres away from a tower 22\displaystyle 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.

    Matches the book, not yet reviewed

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    NCERT’s answer
    \(\displaystyle 45^{\circ}\)
    NCERT_Solution_Class10_Maths_Exemplar_Ch8_Ex8-3_Q14 \[MT=BT-FE=22-1.5=20.5\text{ m},\quad EM=BF=20.5\text{ m} \] \[\tan(\angle TEM)=\frac{MT}{EM}=\frac{20.5}{20.5}=1 \] \[\angle TEM=45^\circ \] Answer: \(\displaystyle 45^\circ\)

  5. Exercise 15

    Show that tan⁡4θ+tan⁡2θ=sec⁡4θ−sec⁡2θ\displaystyle \tan ^4 \theta+\tan ^2 \theta=\sec ^4 \theta-\sec ^2 \theta.

    Not cross-checked

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

    \[\tan^4\theta+\tan^2\theta=\tan^2\theta(\tan^2\theta+1)=\tan^2\theta\sec^2\theta \] \[\sec^4\theta-\sec^2\theta=\sec^2\theta(\sec^2\theta-1)=\sec^2\theta\tan^2\theta \] \[\therefore\ \tan^4\theta+\tan^2\theta=\sec^4\theta-\sec^2\theta \] Answer: Proved.