SolveItClass 10 · NCERT

NCERT Solutions · Class 10 Mathematics Introduction to Trigonometry

19 questions · 19 still being checked

EXERCISE 8.2 1–4 (part 2 of 3)

  1. Exercise 1

    Evaluate the following :
    (i)
    sin⁡60∘cos⁡30∘+sin⁡30∘cos⁡60∘\displaystyle \sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}
    (ii)
    2tan⁡245∘+cos⁡230∘−sin⁡260∘\displaystyle 2 \tan ^{2} 45^{\circ}+\cos ^{2} 30^{\circ}-\sin ^{2} 60^{\circ}
    (iii)
    cos⁡45∘sec⁡30∘+cosec⁡30∘\displaystyle \frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}
    (iv)
    sin⁡30∘+tan⁡45∘−cosec⁡60∘sec⁡30∘+cos⁡60∘+cot⁡45∘\displaystyle \frac{\sin 30^{\circ}+\tan 45^{\circ}-\operatorname{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}
    (v)
    5cos⁡260∘+4sec⁡230∘−tan⁡245∘sin⁡230∘+cos⁡230∘\displaystyle \frac{5 \cos ^{2} 60^{\circ}+4 \sec ^{2} 30^{\circ}-\tan ^{2} 45^{\circ}}{\sin ^{2} 30^{\circ}+\cos ^{2} 30^{\circ}}
    NCERT’s answer
    (i)
    $\displaystyle 1$ (ii) $\displaystyle 2$ (iii) $\displaystyle \frac{3 \sqrt{2}-\sqrt{6}}{8}$ (iv) $\displaystyle \frac{43-24 \sqrt{3}}{11}$ (v) $\displaystyle \frac{67}{12}$

    Working being prepared

  2. Exercise 2

    Choose the correct option and justify your choice :
    (i)
    2tan⁡30∘1+tan⁡230∘=\displaystyle \frac{2 \tan 30^{\circ}}{1+\tan ^{2} 30^{\circ}}= (A) sin⁡60∘\displaystyle \sin 60^{\circ} (B) cos⁡60∘\displaystyle \cos 60^{\circ} (C) tan⁡60∘\displaystyle \tan 60^{\circ} (D) sin⁡30∘\displaystyle \sin 30^{\circ}
    (ii)
    1−tan⁡245∘1+tan⁡245∘=\displaystyle \frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}= (A) tan⁡90∘\displaystyle \tan 90^{\circ} (B) 1\displaystyle 1 (C) sin⁡45∘\displaystyle \sin 45^{\circ} (D) 0\displaystyle 0
    (iii)
    sin⁡2 A=2sin⁡ A\displaystyle \sin 2 \mathrm{~A}=2 \sin \mathrm{~A} is true when A=\displaystyle \mathrm{A}= (A) 0\displaystyle 0° (B) 30\displaystyle 30° (C) 45\displaystyle 45° (D) 60\displaystyle 60°
    (iv)
    2tan⁡30∘1−tan⁡230∘=\displaystyle \frac{2 \tan 30^{\circ}}{1-\tan ^{2} 30^{\circ}}= (A) cos⁡60∘\displaystyle \cos 60^{\circ} (B) sin⁡60∘\displaystyle \sin 60^{\circ} (C) tan⁡60∘\displaystyle \tan 60^{\circ} (D) sin⁡30∘\displaystyle \sin 30^{\circ}
    NCERT’s answer
    (i)
    A (ii) D (iii) A (iv) C

    Working being prepared

  3. Exercise 3

    If tan⁡(A+B)=3\displaystyle \tan (\mathrm{A}+\mathrm{B})=\sqrt{3} and tan⁡(A−B)=13;0∘<A+B≤90∘;A>B\displaystyle \tan (\mathrm{A}-\mathrm{B})=\frac{1}{\sqrt{3}} ; 0^{\circ}<\mathrm{A}+\mathrm{B} \leq 90^{\circ} ; \mathrm{A}>\mathrm{B}, find A and B .
    NCERT’s answer
    $\displaystyle \angle \mathrm{A}=45^{\circ}, \angle \mathrm{B}=15^{\circ}$

    Working being prepared

  4. Exercise 4

    State whether the following are true or false. Justify your answer.
    (i)
    sin⁡(A+B)=sin⁡A+sin⁡B\displaystyle \sin (\mathrm{A}+\mathrm{B})=\sin \mathrm{A}+\sin \mathrm{B}.
    (ii)
    The value of sin⁡θ\displaystyle \sin \theta increases as θ\displaystyle \theta increases.
    (iii)
    The value of cos⁡θ\displaystyle \cos \theta increases as θ\displaystyle \theta increases.
    (iv)
    sin⁡θ=cos⁡θ\displaystyle \sin \theta=\cos \theta for all values of θ\displaystyle \theta.
    (v)
    cot⁡A\displaystyle \cot \mathrm{A} is not defined for A=0∘\displaystyle \mathrm{A}=0^{\circ}.
    NCERT’s answer
    (i)
    False (ii) True (iii) False (iv) False (v) True

    Working being prepared