Search

Samacheer Kalvi 10th Maths Exercise 6.1 Solutions

10th Maths Exercise 6.1 solutions (Chapter 6 Trigonometry) with step-by-step answers for all 19 questions, as per the 2026-27 Samacheer Kalvi textbook.

Question 1

(i)

Prove the following identity. cot⁡θ+tan⁡θ=sec⁡θ cosec θ\cot\theta + \tan\theta = \sec\theta\,\mathrm{cosec}\,\theta

  1. cot⁡θ+tan⁡θ=cos⁡θsin⁡θ+sin⁡θcos⁡θ\cot\theta+\tan\theta=\dfrac{\cos\theta}{\sin\theta}+\dfrac{\sin\theta}{\cos\theta}

  2. =cos⁡2θ+sin⁡2θsin⁡θcos⁡θ=\dfrac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}

  3. =1sin⁡θcos⁡θ=\dfrac{1}{\sin\theta\cos\theta} (since sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1)

  4. =1cos⁡θ×1sin⁡θ=sec⁡θ cosec θ=\dfrac{1}{\cos\theta}\times\dfrac{1}{\sin\theta}=\sec\theta\,\mathrm{cosec}\,\theta

Answer

cot⁡θ+tan⁡θ=sec⁡θ cosec θ\cot\theta+\tan\theta=\sec\theta\,\mathrm{cosec}\,\theta (proved)

(ii)

Prove the following identity. tan⁡4θ+tan⁡2θ=sec⁡4θ−sec⁡2θ\tan^4\theta + \tan^2\theta = \sec^4\theta - \sec^2\theta

  1. tan⁡4θ+tan⁡2θ=tan⁡2θ(tan⁡2θ+1)\tan^4\theta+\tan^2\theta=\tan^2\theta(\tan^2\theta+1)

  2. =(sec⁡2θ−1)(sec⁡2θ)=(\sec^2\theta-1)(\sec^2\theta) (since tan⁡2θ=sec⁡2θ−1\tan^2\theta=\sec^2\theta-1, tan⁡2θ+1=sec⁡2θ\tan^2\theta+1=\sec^2\theta)

  3. =sec⁡4θ−sec⁡2θ=\sec^4\theta-\sec^2\theta

Answer

tan⁡4θ+tan⁡2θ=sec⁡4θ−sec⁡2θ\tan^4\theta+\tan^2\theta=\sec^4\theta-\sec^2\theta (proved)

Question 2

(i)

Prove the following identity. 1−tan⁡2θcot⁡2θ−1=tan⁡2θ\dfrac{1-\tan^2\theta}{\cot^2\theta-1} = \tan^2\theta

  1. 1−tan⁡2θcot⁡2θ−1=1−tan⁡2θ1tan⁡2θ−1\dfrac{1-\tan^2\theta}{\cot^2\theta-1}=\dfrac{1-\tan^2\theta}{\dfrac{1}{\tan^2\theta}-1}

  2. =(1−tan⁡2θ)tan⁡2θ1−tan⁡2θ=\dfrac{(1-\tan^2\theta)\tan^2\theta}{1-\tan^2\theta}

  3. =tan⁡2θ=\tan^2\theta

Answer

1−tan⁡2θcot⁡2θ−1=tan⁡2θ\dfrac{1-\tan^2\theta}{\cot^2\theta-1}=\tan^2\theta (proved)

(ii)

Prove the following identity. cos⁡θ1+sin⁡θ=sec⁡θ−tan⁡θ\dfrac{\cos\theta}{1+\sin\theta} = \sec\theta - \tan\theta

  1. cos⁡θ1+sin⁡θ=cos⁡θ1+sin⁡θ×1−sin⁡θ1−sin⁡θ\dfrac{\cos\theta}{1+\sin\theta}=\dfrac{\cos\theta}{1+\sin\theta}\times\dfrac{1-\sin\theta}{1-\sin\theta}

  2. =cos⁡θ(1−sin⁡θ)1−sin⁡2θ=cos⁡θ(1−sin⁡θ)cos⁡2θ=\dfrac{\cos\theta(1-\sin\theta)}{1-\sin^2\theta}=\dfrac{\cos\theta(1-\sin\theta)}{\cos^2\theta}

  3. =1−sin⁡θcos⁡θ=1cos⁡θ−sin⁡θcos⁡θ=\dfrac{1-\sin\theta}{\cos\theta}=\dfrac{1}{\cos\theta}-\dfrac{\sin\theta}{\cos\theta}

  4. =sec⁡θ−tan⁡θ=\sec\theta-\tan\theta

Answer

cos⁡θ1+sin⁡θ=sec⁡θ−tan⁡θ\dfrac{\cos\theta}{1+\sin\theta}=\sec\theta-\tan\theta (proved)

Question 3

(i)

Prove the following identity. 1+sin⁡θ1−sin⁡θ=sec⁡θ+tan⁡θ\sqrt{\dfrac{1+\sin\theta}{1-\sin\theta}} = \sec\theta + \tan\theta

  1. 1+sin⁡θ1−sin⁡θ=1+sin⁡θ1−sin⁡θ×1+sin⁡θ1+sin⁡θ\sqrt{\dfrac{1+\sin\theta}{1-\sin\theta}}=\sqrt{\dfrac{1+\sin\theta}{1-\sin\theta}\times\dfrac{1+\sin\theta}{1+\sin\theta}}

  2. =(1+sin⁡θ)21−sin⁡2θ=(1+sin⁡θ)2cos⁡2θ=\sqrt{\dfrac{(1+\sin\theta)^2}{1-\sin^2\theta}}=\sqrt{\dfrac{(1+\sin\theta)^2}{\cos^2\theta}}

  3. =1+sin⁡θcos⁡θ=1cos⁡θ+sin⁡θcos⁡θ=\dfrac{1+\sin\theta}{\cos\theta}=\dfrac{1}{\cos\theta}+\dfrac{\sin\theta}{\cos\theta}

  4. =sec⁡θ+tan⁡θ=\sec\theta+\tan\theta

Answer

1+sin⁡θ1−sin⁡θ=sec⁡θ+tan⁡θ\sqrt{\dfrac{1+\sin\theta}{1-\sin\theta}}=\sec\theta+\tan\theta (proved)

(ii)

Prove the following identity. 1+sin⁡θ1−sin⁡θ+1−sin⁡θ1+sin⁡θ=2sec⁡θ\sqrt{\dfrac{1+\sin\theta}{1-\sin\theta}} + \sqrt{\dfrac{1-\sin\theta}{1+\sin\theta}} = 2\sec\theta

  1. 1+sin⁡θ1−sin⁡θ=(1+sin⁡θ)2cos⁡2θ=1+sin⁡θcos⁡θ\sqrt{\dfrac{1+\sin\theta}{1-\sin\theta}}=\sqrt{\dfrac{(1+\sin\theta)^2}{\cos^2\theta}}=\dfrac{1+\sin\theta}{\cos\theta}

  2. 1−sin⁡θ1+sin⁡θ=(1−sin⁡θ)2cos⁡2θ=1−sin⁡θcos⁡θ\sqrt{\dfrac{1-\sin\theta}{1+\sin\theta}}=\sqrt{\dfrac{(1-\sin\theta)^2}{\cos^2\theta}}=\dfrac{1-\sin\theta}{\cos\theta}

  3. Adding, =(1+sin⁡θ)+(1−sin⁡θ)cos⁡θ=2cos⁡θ=\dfrac{(1+\sin\theta)+(1-\sin\theta)}{\cos\theta}=\dfrac{2}{\cos\theta}

  4. =2sec⁡θ=2\sec\theta

Answer

1+sin⁡θ1−sin⁡θ+1−sin⁡θ1+sin⁡θ=2sec⁡θ\sqrt{\dfrac{1+\sin\theta}{1-\sin\theta}}+\sqrt{\dfrac{1-\sin\theta}{1+\sin\theta}}=2\sec\theta (proved)

Question 4

(i)

Prove the following identity. sec⁡6θ=tan⁡6θ+3tan⁡2θsec⁡2θ+1\sec^6\theta = \tan^6\theta + 3\tan^2\theta\sec^2\theta + 1

  1. Take RHS =tan⁡6θ+3tan⁡2θsec⁡2θ+1=\tan^6\theta+3\tan^2\theta\sec^2\theta+1

  2. sec⁡2θ=1+tan⁡2θ\sec^2\theta=1+\tan^2\theta, so 3tan⁡2θsec⁡2θ=3tan⁡2θ(1+tan⁡2θ)=3tan⁡2θ+3tan⁡4θ3\tan^2\theta\sec^2\theta=3\tan^2\theta(1+\tan^2\theta)=3\tan^2\theta+3\tan^4\theta

  3. RHS =1+3tan⁡2θ+3tan⁡4θ+tan⁡6θ=1+3\tan^2\theta+3\tan^4\theta+\tan^6\theta

  4. =(1+tan⁡2θ)3=(1+\tan^2\theta)^3 (since (a+b)3=a3+3a2b+3ab2+b3(a+b)^3=a^3+3a^2b+3ab^2+b^3)

  5. =(sec⁡2θ)3=sec⁡6θ==(\sec^2\theta)^3=\sec^6\theta= LHS

Answer

sec⁡6θ=tan⁡6θ+3tan⁡2θsec⁡2θ+1\sec^6\theta=\tan^6\theta+3\tan^2\theta\sec^2\theta+1 (proved)

(ii)

Prove the following identity. (sin⁡θ+sec⁡θ)2+(cos⁡θ+cosec θ)2=1+(sec⁡θ+cosec θ)2(\sin\theta + \sec\theta)^2 + (\cos\theta + \mathrm{cosec}\,\theta)^2 = 1 + (\sec\theta + \mathrm{cosec}\,\theta)^2

Show steps (6)

  1. LHS =(sin⁡θ+sec⁡θ)2+(cos⁡θ+cosec θ)2=(\sin\theta+\sec\theta)^2+(\cos\theta+\mathrm{cosec}\,\theta)^2

  2. =sin⁡2θ+2sin⁡θsec⁡θ+sec⁡2θ+cos⁡2θ+2cos⁡θ cosec θ+cosec2θ=\sin^2\theta+2\sin\theta\sec\theta+\sec^2\theta+\cos^2\theta+2\cos\theta\,\mathrm{cosec}\,\theta+\mathrm{cosec}^2\theta

  3. =1+2sin⁡θcos⁡θ+2cos⁡θsin⁡θ+sec⁡2θ+cosec2θ=1+2\dfrac{\sin\theta}{\cos\theta}+2\dfrac{\cos\theta}{\sin\theta}+\sec^2\theta+\mathrm{cosec}^2\theta

  4. =1+2(tan⁡θ+cot⁡θ)+sec⁡2θ+cosec2θ=1+2(\tan\theta+\cot\theta)+\sec^2\theta+\mathrm{cosec}^2\theta

  5. tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ=sec⁡θ cosec θ\tan\theta+\cot\theta=\dfrac{1}{\sin\theta\cos\theta}=\sec\theta\,\mathrm{cosec}\,\theta

  6. LHS =1+sec⁡2θ+2sec⁡θ cosec θ+cosec2θ=1+(sec⁡θ+cosec θ)2==1+\sec^2\theta+2\sec\theta\,\mathrm{cosec}\,\theta+\mathrm{cosec}^2\theta=1+(\sec\theta+\mathrm{cosec}\,\theta)^2= RHS

Answer

(sin⁡θ+sec⁡θ)2+(cos⁡θ+cosec θ)2=1+(sec⁡θ+cosec θ)2(\sin\theta+\sec\theta)^2+(\cos\theta+\mathrm{cosec}\,\theta)^2=1+(\sec\theta+\mathrm{cosec}\,\theta)^2 (proved)

Question 5

(i)

Prove the following identity. sec⁡4θ(1−sin⁡4θ)−2tan⁡2θ=1\sec^4\theta(1-\sin^4\theta) - 2\tan^2\theta = 1

  1. 1−sin⁡4θ=(1−sin⁡2θ)(1+sin⁡2θ)=cos⁡2θ(1+sin⁡2θ)1-\sin^4\theta=(1-\sin^2\theta)(1+\sin^2\theta)=\cos^2\theta(1+\sin^2\theta)

  2. LHS =sec⁡4θ⋅cos⁡2θ(1+sin⁡2θ)−2tan⁡2θ=sec⁡2θ(1+sin⁡2θ)−2tan⁡2θ=\sec^4\theta\cdot\cos^2\theta(1+\sin^2\theta)-2\tan^2\theta=\sec^2\theta(1+\sin^2\theta)-2\tan^2\theta

  3. =sec⁡2θ+sin⁡2θcos⁡2θ−2tan⁡2θ=sec⁡2θ+tan⁡2θ−2tan⁡2θ=\sec^2\theta+\dfrac{\sin^2\theta}{\cos^2\theta}-2\tan^2\theta=\sec^2\theta+\tan^2\theta-2\tan^2\theta

  4. =sec⁡2θ−tan⁡2θ=1=\sec^2\theta-\tan^2\theta=1

Answer

sec⁡4θ(1−sin⁡4θ)−2tan⁡2θ=1\sec^4\theta(1-\sin^4\theta)-2\tan^2\theta=1 (proved)

(ii)

Prove the following identity. cot⁡θ−cos⁡θcot⁡θ+cos⁡θ=cosec θ−1cosec θ+1\dfrac{\cot\theta - \cos\theta}{\cot\theta + \cos\theta} = \dfrac{\mathrm{cosec}\,\theta - 1}{\mathrm{cosec}\,\theta + 1}

  1. LHS =cos⁡θsin⁡θ−cos⁡θcos⁡θsin⁡θ+cos⁡θ=\dfrac{\dfrac{\cos\theta}{\sin\theta}-\cos\theta}{\dfrac{\cos\theta}{\sin\theta}+\cos\theta}

  2. =cos⁡θ(1sin⁡θ−1)cos⁡θ(1sin⁡θ+1)=\dfrac{\cos\theta\left(\dfrac{1}{\sin\theta}-1\right)}{\cos\theta\left(\dfrac{1}{\sin\theta}+1\right)}

  3. =cosec θ−1cosec θ+1==\dfrac{\mathrm{cosec}\,\theta-1}{\mathrm{cosec}\,\theta+1}= RHS

Answer

cot⁡θ−cos⁡θcot⁡θ+cos⁡θ=cosec θ−1cosec θ+1\dfrac{\cot\theta-\cos\theta}{\cot\theta+\cos\theta}=\dfrac{\mathrm{cosec}\,\theta-1}{\mathrm{cosec}\,\theta+1} (proved)

Question 6

(i)

Prove the following identity. sin⁡A−sin⁡Bcos⁡A+cos⁡B+cos⁡A−cos⁡Bsin⁡A+sin⁡B=0\dfrac{\sin A - \sin B}{\cos A + \cos B} + \dfrac{\cos A - \cos B}{\sin A + \sin B} = 0

  1. LHS =(sin⁡A−sin⁡B)(sin⁡A+sin⁡B)+(cos⁡A−cos⁡B)(cos⁡A+cos⁡B)(cos⁡A+cos⁡B)(sin⁡A+sin⁡B)=\dfrac{(\sin A-\sin B)(\sin A+\sin B)+(\cos A-\cos B)(\cos A+\cos B)}{(\cos A+\cos B)(\sin A+\sin B)}

  2. Numerator =(sin⁡2A−sin⁡2B)+(cos⁡2A−cos⁡2B)=(\sin^2A-\sin^2B)+(\cos^2A-\cos^2B)

  3. =(sin⁡2A+cos⁡2A)−(sin⁡2B+cos⁡2B)=1−1=0=(\sin^2A+\cos^2A)-(\sin^2B+\cos^2B)=1-1=0

  4. LHS =0(cos⁡A+cos⁡B)(sin⁡A+sin⁡B)=0=\dfrac{0}{(\cos A+\cos B)(\sin A+\sin B)}=0

Answer

sin⁡A−sin⁡Bcos⁡A+cos⁡B+cos⁡A−cos⁡Bsin⁡A+sin⁡B=0\dfrac{\sin A-\sin B}{\cos A+\cos B}+\dfrac{\cos A-\cos B}{\sin A+\sin B}=0 (proved)

(ii)

Prove the following identity. sin⁡3A+cos⁡3Asin⁡A+cos⁡A+sin⁡3A−cos⁡3Asin⁡A−cos⁡A=2\dfrac{\sin^3 A + \cos^3 A}{\sin A + \cos A} + \dfrac{\sin^3 A - \cos^3 A}{\sin A - \cos A} = 2

  1. sin⁡3A+cos⁡3Asin⁡A+cos⁡A=(sin⁡A+cos⁡A)(sin⁡2A−sin⁡Acos⁡A+cos⁡2A)sin⁡A+cos⁡A=1−sin⁡Acos⁡A\dfrac{\sin^3A+\cos^3A}{\sin A+\cos A}=\dfrac{(\sin A+\cos A)(\sin^2A-\sin A\cos A+\cos^2A)}{\sin A+\cos A}=1-\sin A\cos A

  2. sin⁡3A−cos⁡3Asin⁡A−cos⁡A=(sin⁡A−cos⁡A)(sin⁡2A+sin⁡Acos⁡A+cos⁡2A)sin⁡A−cos⁡A=1+sin⁡Acos⁡A\dfrac{\sin^3A-\cos^3A}{\sin A-\cos A}=\dfrac{(\sin A-\cos A)(\sin^2A+\sin A\cos A+\cos^2A)}{\sin A-\cos A}=1+\sin A\cos A

  3. Adding, (1−sin⁡Acos⁡A)+(1+sin⁡Acos⁡A)=2(1-\sin A\cos A)+(1+\sin A\cos A)=2

Answer

sin⁡3A+cos⁡3Asin⁡A+cos⁡A+sin⁡3A−cos⁡3Asin⁡A−cos⁡A=2\dfrac{\sin^3A+\cos^3A}{\sin A+\cos A}+\dfrac{\sin^3A-\cos^3A}{\sin A-\cos A}=2 (proved)

Question 7

(i)

If sin⁡θ+cos⁡θ=3\sin\theta + \cos\theta = \sqrt{3}, then prove that tan⁡θ+cot⁡θ=1\tan\theta + \cot\theta = 1.

  1. sin⁡θ+cos⁡θ=3\sin\theta+\cos\theta=\sqrt3. Squaring both sides,

  2. sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=3\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=3

  3. 1+2sin⁡θcos⁡θ=3⇒sin⁡θcos⁡θ=11+2\sin\theta\cos\theta=3\Rightarrow\sin\theta\cos\theta=1

  4. tan⁡θ+cot⁡θ=sin⁡θcos⁡θ+cos⁡θsin⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ\tan\theta+\cot\theta=\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}=\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\dfrac{1}{\sin\theta\cos\theta}

  5. =11=1=\dfrac{1}{1}=1

Answer

tan⁡θ+cot⁡θ=1\tan\theta+\cot\theta=1 (proved)

(ii)

If 3sin⁡θ−cos⁡θ=0\sqrt{3}\sin\theta - \cos\theta = 0, then show that tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta = \dfrac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}

Show steps (6)

  1. 3sin⁡θ−cos⁡θ=0⇒3sin⁡θ=cos⁡θ\sqrt3\sin\theta-\cos\theta=0\Rightarrow\sqrt3\sin\theta=\cos\theta

  2. tan⁡θ=13⇒θ=30∘\tan\theta=\dfrac{1}{\sqrt3}\Rightarrow\theta=30^\circ

  3. LHS =tan⁡3θ=tan⁡90∘=\tan3\theta=\tan90^\circ, which is undefined (infinite).

  4. RHS: denominator 1−3tan⁡2θ=1−3(13)=01-3\tan^2\theta=1-3\left(\dfrac13\right)=0

  5. Numerator =3(13)−(13)3=833≠0=3\left(\dfrac{1}{\sqrt3}\right)-\left(\dfrac{1}{\sqrt3}\right)^3=\dfrac{8}{3\sqrt3}\neq0, so RHS is also undefined (infinite).

  6. Hence LHS == RHS.

Answer

θ=30∘\theta=30^\circ and both sides equal tan⁡90∘\tan 90^\circ, so tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan3\theta=\dfrac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta} (proved)

Question 8

(i)

If cos⁡αcos⁡β=m\dfrac{\cos\alpha}{\cos\beta} = m and cos⁡αsin⁡β=n\dfrac{\cos\alpha}{\sin\beta} = n, then prove that (m2+n2)cos⁡2β=n2(m^2 + n^2)\cos^2\beta = n^2

  1. m=cos⁡αcos⁡β,  n=cos⁡αsin⁡βm=\dfrac{\cos\alpha}{\cos\beta},\; n=\dfrac{\cos\alpha}{\sin\beta}

  2. (m2+n2)cos⁡2β=(cos⁡2αcos⁡2β+cos⁡2αsin⁡2β)cos⁡2β(m^2+n^2)\cos^2\beta=\left(\dfrac{\cos^2\alpha}{\cos^2\beta}+\dfrac{\cos^2\alpha}{\sin^2\beta}\right)\cos^2\beta

  3. =cos⁡2α+cos⁡2αcos⁡2βsin⁡2β=cos⁡2α(sin⁡2β+cos⁡2β)sin⁡2β=\cos^2\alpha+\dfrac{\cos^2\alpha\cos^2\beta}{\sin^2\beta}=\dfrac{\cos^2\alpha(\sin^2\beta+\cos^2\beta)}{\sin^2\beta}

  4. =cos⁡2αsin⁡2β=(cos⁡αsin⁡β)2=n2=\dfrac{\cos^2\alpha}{\sin^2\beta}=\left(\dfrac{\cos\alpha}{\sin\beta}\right)^2=n^2

Answer

(m2+n2)cos⁡2β=n2(m^2+n^2)\cos^2\beta=n^2 (proved)

(ii)

If cot⁡θ+tan⁡θ=x\cot\theta + \tan\theta = x and sec⁡θ−cos⁡θ=y\sec\theta - \cos\theta = y, then prove that (x2y)23−(xy2)23=1(x^2y)^{\frac{2}{3}} - (xy^2)^{\frac{2}{3}} = 1

  1. x=cot⁡θ+tan⁡θ=cos⁡2θ+sin⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θx=\cot\theta+\tan\theta=\dfrac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}=\dfrac{1}{\sin\theta\cos\theta}

  2. y=sec⁡θ−cos⁡θ=1−cos⁡2θcos⁡θ=sin⁡2θcos⁡θy=\sec\theta-\cos\theta=\dfrac{1-\cos^2\theta}{\cos\theta}=\dfrac{\sin^2\theta}{\cos\theta}

  3. x2y=1sin⁡2θcos⁡2θ×sin⁡2θcos⁡θ=1cos⁡3θx^2y=\dfrac{1}{\sin^2\theta\cos^2\theta}\times\dfrac{\sin^2\theta}{\cos\theta}=\dfrac{1}{\cos^3\theta}, so (x2y)23=1cos⁡2θ=sec⁡2θ(x^2y)^{\frac23}=\dfrac{1}{\cos^2\theta}=\sec^2\theta

  4. xy2=1sin⁡θcos⁡θ×sin⁡4θcos⁡2θ=sin⁡3θcos⁡3θ=tan⁡3θxy^2=\dfrac{1}{\sin\theta\cos\theta}\times\dfrac{\sin^4\theta}{\cos^2\theta}=\dfrac{\sin^3\theta}{\cos^3\theta}=\tan^3\theta, so (xy2)23=tan⁡2θ(xy^2)^{\frac23}=\tan^2\theta

  5. (x2y)23−(xy2)23=sec⁡2θ−tan⁡2θ=1(x^2y)^{\frac23}-(xy^2)^{\frac23}=\sec^2\theta-\tan^2\theta=1

Answer

(x2y)23−(xy2)23=1(x^2y)^{\frac23}-(xy^2)^{\frac23}=1 (proved)

Question 9

(i)

If sin⁡θ+cos⁡θ=p\sin\theta + \cos\theta = p and sec⁡θ+cosec θ=q\sec\theta + \mathrm{cosec}\,\theta = q, then prove that q(p2−1)=2pq(p^2 - 1) = 2p

  1. p=sin⁡θ+cos⁡θp=\sin\theta+\cos\theta, so p2=1+2sin⁡θcos⁡θp^2=1+2\sin\theta\cos\theta, giving p2−1=2sin⁡θcos⁡θp^2-1=2\sin\theta\cos\theta

  2. q=sec⁡θ+cosec θ=1cos⁡θ+1sin⁡θ=sin⁡θ+cos⁡θsin⁡θcos⁡θ=psin⁡θcos⁡θq=\sec\theta+\mathrm{cosec}\,\theta=\dfrac{1}{\cos\theta}+\dfrac{1}{\sin\theta}=\dfrac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=\dfrac{p}{\sin\theta\cos\theta}

  3. q(p2−1)=psin⁡θcos⁡θ×2sin⁡θcos⁡θq(p^2-1)=\dfrac{p}{\sin\theta\cos\theta}\times2\sin\theta\cos\theta

  4. =2p=2p

Answer

q(p2−1)=2pq(p^2-1)=2p (proved)

(ii)

If sin⁡θ(1+sin⁡2θ)=cos⁡2θ\sin\theta(1 + \sin^2\theta) = \cos^2\theta, then prove that cos⁡6θ−4cos⁡4θ+8cos⁡2θ=4\cos^6\theta - 4\cos^4\theta + 8\cos^2\theta = 4

Show steps (6)

  1. Given sin⁡θ(1+sin⁡2θ)=cos⁡2θ\sin\theta(1+\sin^2\theta)=\cos^2\theta and sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta

  2. sin⁡θ(1+1−cos⁡2θ)=cos⁡2θ⇒sin⁡θ(2−cos⁡2θ)=cos⁡2θ\sin\theta(1+1-\cos^2\theta)=\cos^2\theta\Rightarrow\sin\theta(2-\cos^2\theta)=\cos^2\theta

  3. Squaring: sin⁡2θ(2−cos⁡2θ)2=cos⁡4θ\sin^2\theta(2-\cos^2\theta)^2=\cos^4\theta

  4. (1−cos⁡2θ)(4−4cos⁡2θ+cos⁡4θ)=cos⁡4θ(1-\cos^2\theta)(4-4\cos^2\theta+\cos^4\theta)=\cos^4\theta

  5. 4−8cos⁡2θ+5cos⁡4θ−cos⁡6θ=cos⁡4θ4-8\cos^2\theta+5\cos^4\theta-\cos^6\theta=\cos^4\theta

  6. cos⁡6θ−4cos⁡4θ+8cos⁡2θ=4\cos^6\theta-4\cos^4\theta+8\cos^2\theta=4

Answer

cos⁡6θ−4cos⁡4θ+8cos⁡2θ=4\cos^6\theta-4\cos^4\theta+8\cos^2\theta=4 (proved)

Q10. If cos⁡θ1+sin⁡θ=1a\dfrac{\cos\theta}{1+\sin\theta} = \dfrac{1}{a} , then prove that a2−1a2+1=sin⁡θ\dfrac{a^2-1}{a^2+1} = \sin\theta

  1. cos⁡θ1+sin⁡θ=1a⇒a=1+sin⁡θcos⁡θ\dfrac{\cos\theta}{1+\sin\theta}=\dfrac1a\Rightarrow a=\dfrac{1+\sin\theta}{\cos\theta}

  2. a2=(1+sin⁡θ)2cos⁡2θ=(1+sin⁡θ)2(1+sin⁡θ)(1−sin⁡θ)=1+sin⁡θ1−sin⁡θa^2=\dfrac{(1+\sin\theta)^2}{\cos^2\theta}=\dfrac{(1+\sin\theta)^2}{(1+\sin\theta)(1-\sin\theta)}=\dfrac{1+\sin\theta}{1-\sin\theta}

  3. a2−1a2+1=1+sin⁡θ1−sin⁡θ−11+sin⁡θ1−sin⁡θ+1=(1+sin⁡θ)−(1−sin⁡θ)(1+sin⁡θ)+(1−sin⁡θ)\dfrac{a^2-1}{a^2+1}=\dfrac{\dfrac{1+\sin\theta}{1-\sin\theta}-1}{\dfrac{1+\sin\theta}{1-\sin\theta}+1}=\dfrac{(1+\sin\theta)-(1-\sin\theta)}{(1+\sin\theta)+(1-\sin\theta)}

  4. =2sin⁡θ2=sin⁡θ=\dfrac{2\sin\theta}{2}=\sin\theta

Answer

a2−1a2+1=sin⁡θ\dfrac{a^2-1}{a^2+1}=\sin\theta (proved)

Quick check

  1. A tower is 60 m heigh. Its shadow reduces by xx metres when the angle of elevation of the sun increases from 30∘30^\circ to 45∘45^\circ then xx is equal to

    1. (1) 41.92 m
    2. (2) 43.92 m
    3. (3) 43 m
    4. (4) 45.6 m
    Show answer

    Answer: (2) 43.92 m

    1. Shadow at 30∘30^\circ: 60tan⁡30∘=603\dfrac{60}{\tan 30^\circ}=60\sqrt3 m.

    2. Shadow at 45∘45^\circ: 60tan⁡45∘=60\dfrac{60}{\tan 45^\circ}=60 m.

    3. x=603−60=60(1.732−1)=43.92x=60\sqrt3-60=60(1.732-1)=43.92 m.

    Question 1 of 5

  2. If sin⁡θ+cos⁡θ=a\sin\theta + \cos\theta = a and sec⁡θ+cosec θ=b\sec\theta + \mathrm{cosec}\,\theta = b, then the value of b(a2−1)b(a^2-1) is equal to

    1. (1) 2a2a
    2. (2) 3a3a
    3. (3) 00
    4. (4) 2ab2ab
    Show answer

    Answer: (1) 2a2a

    1. a2−1=(sin⁡θ+cos⁡θ)2−1=2sin⁡θcos⁡θa^2-1=(\sin\theta+\cos\theta)^2-1=2\sin\theta\cos\theta.

    2. b=sec⁡θ+cosec θ=sin⁡θ+cos⁡θsin⁡θcos⁡θ=asin⁡θcos⁡θb=\sec\theta+\mathrm{cosec}\,\theta=\dfrac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=\dfrac{a}{\sin\theta\cos\theta}.

    3. b(a2−1)=asin⁡θcos⁡θ×2sin⁡θcos⁡θ=2ab(a^2-1)=\dfrac{a}{\sin\theta\cos\theta}\times 2\sin\theta\cos\theta=2a.

    Question 2 of 5

  3. acot⁡θ+b cosec θ=pa\cot\theta + b\,\mathrm{cosec}\,\theta = p and bcot⁡θ+a cosec θ=qb\cot\theta + a\,\mathrm{cosec}\,\theta = q then p2−q2p^2 - q^2 is equal to

    1. (1) a2−b2a^2 - b^2
    2. (2) b2−a2b^2 - a^2
    3. (3) a2+b2a^2 + b^2
    4. (4) b−ab - a
    Show answer

    Answer: (2) b2−a2b^2 - a^2

    1. p2−q2=(acot⁡θ+b cosec θ)2−(bcot⁡θ+a cosec θ)2p^2-q^2=(a\cot\theta+b\,\mathrm{cosec}\,\theta)^2-(b\cot\theta+a\,\mathrm{cosec}\,\theta)^2.

    2. =(a2−b2)cot⁡2θ−(a2−b2)cosec2θ=(a^2-b^2)\cot^2\theta-(a^2-b^2)\mathrm{cosec}^2\theta (the 2ab2ab terms cancel).

    3. =(a2−b2)(cot⁡2θ−cosec2θ)=(a2−b2)(−1)=b2−a2=(a^2-b^2)(\cot^2\theta-\mathrm{cosec}^2\theta)=(a^2-b^2)(-1)=b^2-a^2.

    Question 3 of 5

  4. If the ratio of the height of a tower and the length of its shadow is 3:1\sqrt{3}:1, then the angle of elevation of the sun has measure

    1. (1) 45∘45^\circ
    2. (2) 30∘30^\circ
    3. (3) 90∘90^\circ
    4. (4) 60∘60^\circ
    Show answer

    Answer: (4) 60∘60^\circ

    1. tan⁡θ=heightshadow=31=3\tan\theta=\dfrac{\text{height}}{\text{shadow}}=\dfrac{\sqrt3}{1}=\sqrt3.

    2. θ=60∘\theta=60^\circ.

    Question 4 of 5

  5. tan⁡θ cosec2θ−tan⁡θ\tan\theta\,\mathrm{cosec}^2\theta - \tan\theta is equal to

    1. (1) sec⁡θ\sec\theta
    2. (2) cot⁡2θ\cot^2\theta
    3. (3) sin⁡θ\sin\theta
    4. (4) cot⁡θ\cot\theta
    Show answer

    Answer: (4) cot⁡θ\cot\theta

    1. tan⁡θ cosec2θ−tan⁡θ=tan⁡θ(cosec2θ−1)\tan\theta\,\mathrm{cosec}^2\theta-\tan\theta=\tan\theta(\mathrm{cosec}^2\theta-1).

    2. cosec2θ−1=cot⁡2θ\mathrm{cosec}^2\theta-1=\cot^2\theta.

    3. =tan⁡θ⋅cot⁡2θ=cot⁡θ=\tan\theta\cdot\cot^2\theta=\cot\theta.

    Question 5 of 5
Chapter 6 formulas
  • Pythagorean identity

    sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

  • Identity with tangent and secant

    1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta

  • Identity with cotangent and cosecant

    1+cot⁡2θ=cosec⁡2θ1 + \cot^2\theta = \operatorname{cosec}^2\theta

  • Trigonometric identity (definition)

    An equation in trigonometric ratios that is true for all values of the angle\text{An equation in trigonometric ratios that is true for all values of the angle}

  • Line of sight

    The line drawn from the eye of an observer to the point on the object being viewed\text{The line drawn from the eye of an observer to the point on the object being viewed}

  • Angle of elevation

    Angle between the line of sight and the horizontal, when the object is above the horizontal level\text{Angle between the line of sight and the horizontal, when the object is above the horizontal level}

  • Angle of depression

    Angle between the line of sight and the horizontal, when the object is below the horizontal level\text{Angle between the line of sight and the horizontal, when the object is below the horizontal level}

  • Use of trigonometric ratios

    Height or length of an object, or distance between two far objects, can be found using trigonometric ratios\text{Height or length of an object, or distance between two far objects, can be found using trigonometric ratios}

Answers checked against the Samacheer Kalvi textbook · Updated

Spotted a mistake? Email us at [email protected] and we will fix it.