SolveItClass 10 · NCERT

NCERT Solutions · Class 10 Mathematics Introduction to Trigonometry

19 questions · 19 still being checked

EXERCISE 8.3 1–4 (part 3 of 3)

  1. Exercise 1

    Express the trigonometric ratios sin⁡A,sec⁡A\displaystyle \sin \mathrm{A}, \sec \mathrm{A} and tan⁡A\displaystyle \tan \mathrm{A} in terms of cot⁡A\displaystyle \cot \mathrm{A}.
    NCERT’s answer
    $\displaystyle \sin \mathrm{A}=\frac{1}{\sqrt{1+\cot ^{2} \mathrm{~A}}}, \tan \mathrm{~A}=\frac{1}{\cot \mathrm{~A}}, \sec \mathrm{~A}=\frac{\sqrt{1+\cot ^{2} \mathrm{~A}}}{\cot \mathrm{~A}}$

    Working being prepared

  2. Exercise 2

    Write all the other trigonometric ratios of ∠A\displaystyle \angle \mathrm{A} in terms of sec⁡A\displaystyle \sec \mathrm{A}.
    NCERT’s answer
    $\displaystyle \sin \mathrm{A}=\frac{\sqrt{\sec ^{2} \mathrm{~A}-1}}{\sec \mathrm{~A}}, \cos \mathrm{~A}=\frac{1}{\sec \mathrm{~A}}, \tan \mathrm{~A}=\sqrt{\sec ^{2} \mathrm{~A}-1}$ \cot \mathrm{A}=\frac{1}{\sqrt{\sec ^{2} \mathrm{~A}-1}}, \operatorname{cosec} \mathrm{~A}=\frac{\sec \mathrm{A}}{\sqrt{\sec ^{2} \mathrm{~A}-1}} $$

    Working being prepared

  3. Exercise 3

    Choose the correct option. Justify your choice.
    (i)
    9sec⁡2 A−9tan⁡2 A=\displaystyle 9 \sec ^{2} \mathrm{~A}-9 \tan ^{2} \mathrm{~A}= (A) 1\displaystyle 1 (B) 9\displaystyle 9 (C) 8\displaystyle 8 (D) 0\displaystyle 0
    (ii)
    (1+tan⁡θ+sec⁡θ)(1+cot⁡θ−cosec⁡θ)=\displaystyle (1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)= (A) 0\displaystyle 0 (B) 1\displaystyle 1 (C) 2\displaystyle 2 (D) -1\displaystyle 1
    (iii)
    (sec⁡A+tan⁡A)(1−sin⁡A)=\displaystyle (\sec \mathrm{A}+\tan \mathrm{A})(1-\sin \mathrm{A})= (A) sec⁡A\displaystyle \sec \mathrm{A} (B) sin⁡A\displaystyle \sin \mathrm{A} (C) cosec⁡A\displaystyle \operatorname{cosec} \mathrm{A} (D) cos⁡A\displaystyle \cos \mathrm{A}
    (iv)
    1+tan⁡2 A1+cot⁡2 A=\displaystyle \frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}= (A) sec⁡2 A\displaystyle \sec ^{2} \mathrm{~A} (B) -1\displaystyle 1 (C) cot⁡2 A\displaystyle \cot ^{2} \mathrm{~A} (D) tan⁡2 A\displaystyle \tan ^{2} \mathrm{~A}
    NCERT’s answer
    (i)
    B (ii) C (iii) D (iv) D

    Working being prepared

  4. Exercise 4

    Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
    (i)
    (cosec⁡θ−cot⁡θ)2=1−cos⁡θ1+cos⁡θ\displaystyle (\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta}
    (ii)
    cos⁡A1+sin⁡A+1+sin⁡Acos⁡A=2sec⁡ A\displaystyle \frac{\cos \mathrm{A}}{1+\sin \mathrm{A}}+\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}}=2 \sec \mathrm{~A}
    (iii)
    tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θcosec⁡θ\displaystyle \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta [Hint : Write the expression in terms of sin⁡θ\displaystyle \sin \theta and cos⁡θ\displaystyle \cos \theta ]
    (iv)
    1+sec⁡Asec⁡A=sin⁡2 A1−cos⁡A\displaystyle \frac{1+\sec \mathrm{A}}{\sec \mathrm{A}}=\frac{\sin ^{2} \mathrm{~A}}{1-\cos \mathrm{A}} [Hint : Simplify LHS and RHS separately]
    (v)
    cos⁡A−sin⁡A+1cos⁡ A+sin⁡A−1=cosec⁡A+cot⁡A\displaystyle \frac{\cos \mathrm{A}-\sin \mathrm{A}+1}{\cos \mathrm{~A}+\sin \mathrm{A}-1}=\operatorname{cosec} \mathrm{A}+\cot \mathrm{A}, using the identity cosec⁡2 A=1+cot⁡2 A\displaystyle \operatorname{cosec}^{2} \mathrm{~A}=1+\cot ^{2} \mathrm{~A}.
    (vi)
    1+sin⁡A1−sin⁡A=sec⁡A+tan⁡A\displaystyle \sqrt{\frac{1+\sin \mathrm{A}}{1-\sin \mathrm{A}}}=\sec \mathrm{A}+\tan \mathrm{A}
    (vii)
    sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\displaystyle \frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta
    (viii)
    (sin⁡A+cosec⁡A)2+(cos⁡A+sec⁡A)2=7+tan⁡2 A+cot⁡2 A\displaystyle (\sin \mathrm{A}+\operatorname{cosec} \mathrm{A})^{2}+(\cos \mathrm{A}+\sec \mathrm{A})^{2}=7+\tan ^{2} \mathrm{~A}+\cot ^{2} \mathrm{~A}
    (ix)
    (cosec⁡A−sin⁡A)(sec⁡A−cos⁡A)=1tan⁡ A+cot⁡A\displaystyle (\operatorname{cosec} \mathrm{A}-\sin \mathrm{A})(\sec \mathrm{A}-\cos \mathrm{A})=\frac{1}{\tan \mathrm{~A}+\cot \mathrm{A}} [Hint : Simplify LHS and RHS separately]
    (x)
    (1+tan⁡2 A1+cot⁡2 A)=(1−tan⁡A1−cot⁡A)2=tan⁡2 A\displaystyle \left(\frac{1+\tan ^{2} \mathrm{~A}}{1+\cot ^{2} \mathrm{~A}}\right)=\left(\frac{1-\tan \mathrm{A}}{1-\cot \mathrm{A}}\right)^{2}=\tan ^{2} \mathrm{~A}

    Solution being prepared