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NCERT Exemplar · Class 11 Mathematics Trigonometric Functions

76 questions · 76 still being checked

EXERCISE 3.3 1–10 (part 1 of 8)

  1. Exercise 1

    Prove that tan⁡A+sec⁡A−1tan⁡ A−sec⁡A+1=1+sin⁡Acos⁡A\displaystyle \frac{\tan \mathrm{A}+\sec \mathrm{A}-1}{\tan \mathrm{~A}-\sec \mathrm{A}+1}=\frac{1+\sin \mathrm{A}}{\cos \mathrm{A}}

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    Write $\displaystyle 1$ as \(\displaystyle \sec^2A-\tan^2A\) in the numerator.\[\tan A+\sec A-1=(\tan A+\sec A)-(\sec A+\tan A)(\sec A-\tan A) \]\[=(\tan A+\sec A)(1-\sec A+\tan A) \]\[\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{(\tan A+\sec A)(\tan A-\sec A+1)}{\tan A-\sec A+1}=\tan A+\sec A \]\[\tan A+\sec A=\frac{\sin A}{\cos A}+\frac{1}{\cos A}=\frac{1+\sin A}{\cos A} \]Answer: both sides equal \(\displaystyle \dfrac{1+\sin A}{\cos A}\).

  2. Exercise 2

    If 2sin⁡α1+cos⁡α+sin⁡α=y\displaystyle \frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=y, then prove that 1−cos⁡α+sin⁡α1+sin⁡α\displaystyle \frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} is also equal to y\displaystyle y. [Hint:Express 1−cos⁡α+sin⁡α1+sin⁡α=1−cos⁡α+sin⁡α1+sin⁡α⋅1+cos⁡α+sin⁡α1+cos⁡α+sin⁡α\displaystyle \frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}=\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha} \cdot \frac{1+\cos \alpha+\sin \alpha}{1+\cos \alpha+\sin \alpha}]

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    Multiply the numerator and denominator by \(\displaystyle 1+\cos\alpha+\sin\alpha\).\[\frac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}=\frac{(1+\sin\alpha)^2-\cos^2\alpha}{(1+\sin\alpha)(1+\cos\alpha+\sin\alpha)} \]\[(1+\sin\alpha)^2-\cos^2\alpha=1+2\sin\alpha+\sin^2\alpha-\cos^2\alpha=2\sin^2\alpha+2\sin\alpha \]\[=2\sin\alpha\,(1+\sin\alpha) \]\[\frac{2\sin\alpha\,(1+\sin\alpha)}{(1+\sin\alpha)(1+\cos\alpha+\sin\alpha)}=\frac{2\sin\alpha}{1+\cos\alpha+\sin\alpha}=y \]Answer: \(\displaystyle \dfrac{1-\cos\alpha+\sin\alpha}{1+\sin\alpha}=y\).

  3. Exercise 3

    If msin⁡θ=nsin⁡(θ+2α)\displaystyle m \sin \theta=n \sin (\theta+2 \alpha), then prove that tan⁡(θ+α)cot⁡α=m+nm−n\displaystyle \tan (\theta+\alpha) \cot \alpha=\frac{m+n}{m-n} [Hint: Express sin⁡(θ+2α)sin⁡θ=mn\displaystyle \frac{\sin (\theta+2 \alpha)}{\sin \theta}=\frac{m}{n} and apply componendo and dividendo]

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    From the given relation:\[\frac{\sin(\theta+2\alpha)}{\sin\theta}=\frac{m}{n} \]Componendo and dividendo:\[\frac{\sin(\theta+2\alpha)+\sin\theta}{\sin(\theta+2\alpha)-\sin\theta}=\frac{m+n}{m-n} \]Sum-to-product on the left:\[\frac{2\sin(\theta+\alpha)\cos\alpha}{2\cos(\theta+\alpha)\sin\alpha}=\frac{m+n}{m-n} \]\[\tan(\theta+\alpha)\cot\alpha=\frac{m+n}{m-n} \]Answer: \(\displaystyle \tan(\theta+\alpha)\cot\alpha=\dfrac{m+n}{m-n}\).

  4. Exercise 4

    If cos⁡(α+β)=45\displaystyle \cos (\alpha+\beta)=\frac{4}{5} and sin⁡(α−β)=513\displaystyle \sin (\alpha-\beta)=\frac{5}{13}, where α\displaystyle \alpha lie between 0\displaystyle 0 and π4\displaystyle \frac{\pi}{4}, find the value of tan⁡2α\displaystyle \tan 2 \alpha [Hint: Express tan⁡2α\displaystyle \tan 2 \alpha as tan⁡(α+β+α−β]\displaystyle \tan (\alpha+\beta+\alpha-\beta]

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    NCERT’s answer
    \(\displaystyle \frac{56}{33}\)
    Only \(\displaystyle 0<\alpha<\dfrac{\pi}{4}\) is given, so \(\displaystyle 0<2\alpha<\dfrac{\pi}{2}\): both \(\displaystyle \sin2\alpha\) and \(\displaystyle \cos2\alpha\) are positive, and these fix the signs below.\[\sin(\alpha+\beta)=\tfrac35\,s,\qquad \cos(\alpha-\beta)=\tfrac{12}{13}\,t,\qquad s,t=\pm1 \]\[\sin2\alpha=\sin(\alpha+\beta)\cos(\alpha-\beta)+\cos(\alpha+\beta)\sin(\alpha-\beta)=\frac{36\,st+20}{65} \]\[\sin2\alpha>0\ \Rightarrow\ st=1,\qquad \sin2\alpha=\frac{56}{65} \]\[\cos2\alpha=\cos(\alpha+\beta)\cos(\alpha-\beta)-\sin(\alpha+\beta)\sin(\alpha-\beta)=\frac{48\,t-15\,s}{65}=\frac{33\,t}{65} \]\[\cos2\alpha>0\ \Rightarrow\ t=1,\qquad \cos2\alpha=\frac{33}{65} \]\[\tan2\alpha=\frac{56/65}{33/65}=\frac{56}{33} \]Answer: \(\displaystyle \tan2\alpha=\dfrac{56}{33}\).

  5. Exercise 5

    If tan⁡x=ba\displaystyle \tan x=\frac{b}{a}, then find the value of a+ba−b+a−ba+b\displaystyle \sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}

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    NCERT’s answer
    \(\displaystyle \frac{2 \cos x}{\sqrt{\cos 2 x}}\)
    Divide the numerator and denominator inside each root by \(\displaystyle a\).\[\sqrt{\frac{a+b}{a-b}}+\sqrt{\frac{a-b}{a+b}}=\sqrt{\frac{1+\tan x}{1-\tan x}}+\sqrt{\frac{1-\tan x}{1+\tan x}} \]The roots are real only when \(\displaystyle -1<\tan x<1\), so \(\displaystyle 1\pm\tan x>0\):\[=\frac{(1+\tan x)+(1-\tan x)}{\sqrt{(1+\tan x)(1-\tan x)}}=\frac{2}{\sqrt{1-\tan^2x}} \]\[1-\tan^2x=\frac{\cos2x}{\cos^2x}\ \Rightarrow\ \frac{2}{\sqrt{1-\tan^2x}}=\frac{2\cos x}{\sqrt{\cos2x}}\qquad(\cos x>0) \]Answer: \(\displaystyle \dfrac{2}{\sqrt{1-\tan^2x}}=\dfrac{2\cos x}{\sqrt{\cos2x}}\).

  6. Exercise 6

    Prove that cos⁡θcos⁡θ2−cos⁡3θcos⁡9θ2=sin⁡7θsin⁡8θ\displaystyle \cos \theta \cos \frac{\theta}{2}-\cos 3 \theta \cos \frac{9 \theta}{2}=\sin 7 \theta \sin 8 \theta. [Hint: Express L.H.S. =12[2cos⁡θcos⁡θ2−2cos⁡3θcos⁡9θ2]\displaystyle =\frac{1}{2}\left[2 \cos \theta \cos \frac{\theta}{2}-2 \cos 3 \theta \cos \frac{9 \theta}{2}\right]

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    Using \(\displaystyle 2\cos A\cos B=\cos(A+B)+\cos(A-B)\) on each product:\[2\cos\theta\cos\frac{\theta}{2}=\cos\frac{3\theta}{2}+\cos\frac{\theta}{2} \]\[2\cos3\theta\cos\frac{9\theta}{2}=\cos\frac{15\theta}{2}+\cos\frac{3\theta}{2} \]\[\text{L.H.S.}=\frac12\left[\cos\frac{\theta}{2}-\cos\frac{15\theta}{2}\right] \]\[=\frac12\cdot2\sin\frac{\dfrac{\theta}{2}+\dfrac{15\theta}{2}}{2}\,\sin\frac{\dfrac{15\theta}{2}-\dfrac{\theta}{2}}{2}\qquad\left(\cos C-\cos D=2\sin\frac{C+D}{2}\sin\frac{D-C}{2}\right) \]\[=\sin4\theta\,\sin\frac{7\theta}{2} \]This differs from the printed R.H.S., e.g. at \(\displaystyle \theta=\dfrac{\pi}{8}\):\[\sin4\theta\sin\frac{7\theta}{2}=\sin\frac{\pi}{2}\sin\frac{7\pi}{16}\approx0.981,\qquad \sin7\theta\sin8\theta=\sin\frac{7\pi}{8}\sin\pi=0 \]Answer: \(\displaystyle \cos\theta\cos\dfrac{\theta}{2}-\cos3\theta\cos\dfrac{9\theta}{2}=\sin4\theta\sin\dfrac{7\theta}{2}\).NCERT prints: R.H.S. \(\displaystyle =\sin7\theta\sin8\theta\) -- false: that is the true R.H.S. \(\displaystyle \sin\dfrac{7\theta}{2}\sin\dfrac{8\theta}{2}\) with each angle doubled.

  7. Exercise 7

    If acos⁡θ+bsin⁡θ=m\displaystyle a \cos \theta+b \sin \theta=m and asin⁡θ−bcos⁡θ=n\displaystyle a \sin \theta-b \cos \theta=n, then show that a2+b2=m2+n2\displaystyle a^2+b^2=m^2+n^2

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    \[m^2+n^2=(a\cos\theta+b\sin\theta)^2+(a\sin\theta-b\cos\theta)^2 \]\[=a^2\cos^2\theta+2ab\sin\theta\cos\theta+b^2\sin^2\theta+a^2\sin^2\theta-2ab\sin\theta\cos\theta+b^2\cos^2\theta \]\[=a^2(\cos^2\theta+\sin^2\theta)+b^2(\sin^2\theta+\cos^2\theta) \]\[=a^2+b^2 \]Answer: \(\displaystyle m^2+n^2=a^2+b^2\).

  8. Exercise 8

    Find the value of tan⁡22∘30′\displaystyle \tan 22^{\circ} 30^{\prime}. [Hint: Let θ=45∘\displaystyle \theta=45^{\circ}, use tan⁡θ2=sin⁡θ2cos⁡θ2=2sin⁡θ2cos⁡θ22cos⁡2θ2=sin⁡θ1+cos⁡θ\displaystyle \tan \frac{\theta}{2}=\frac{\sin \dfrac{\theta}{2}}{\cos \dfrac{\theta}{2}}=\frac{2 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2}}{2 \cos ^2 \dfrac{\theta}{2}}=\frac{\sin \theta}{1+\cos \theta} ]

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    NCERT’s answer
    \(\displaystyle \frac{1}{\sqrt{2}+1}\)
    Take \(\displaystyle \theta=45^\circ\), so \(\displaystyle \dfrac{\theta}{2}=22^\circ30'\).\[\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta} \]\[\tan22^\circ30'=\frac{\sin45^\circ}{1+\cos45^\circ}=\frac{1/\sqrt2}{1+1/\sqrt2}=\frac{1}{\sqrt2+1} \]\[=\frac{1}{\sqrt2+1}\cdot\frac{\sqrt2-1}{\sqrt2-1}=\sqrt2-1 \]Answer: \(\displaystyle \tan22^\circ30'=\sqrt2-1\).

  9. Exercise 9

    Prove that sin⁡4 A=4sin⁡ Acos⁡3 A−4cos⁡ Asin⁡3 A\displaystyle \sin 4 \mathrm{~A}=4 \sin \mathrm{~A} \cos ^3 \mathrm{~A}-4 \cos \mathrm{~A} \sin ^3 \mathrm{~A}.

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    Write \(\displaystyle 4A=2(2A)\) and apply the double-angle formulas twice.\[\sin 4A = 2\sin 2A\cos 2A \] \[\sin 2A = 2\sin A\cos A,\qquad \cos 2A=\cos^2A-\sin^2A \] \[\sin 4A = 2\,(2\sin A\cos A)(\cos^2A-\sin^2A) \] \[\sin 4A = 4\sin A\cos^3A-4\cos A\sin^3A \]Answer: \(\displaystyle \sin 4A = 4\sin A\cos^3A-4\cos A\sin^3A\)

  10. Exercise 10

    If tan⁡θ+sin⁡θ=m\displaystyle \tan \theta+\sin \theta=m and tan⁡θ−sin⁡θ=n\displaystyle \tan \theta-\sin \theta=n, then prove that m2−n2=4sin⁡θtan⁡θ\displaystyle m^2-n^2=4 \sin \theta \tan \theta [Hint: m+n=2tan⁡θ,m−n=2sin⁡θ\displaystyle m+n=2 \tan \theta, m-n=2 \sin \theta, then use m2−n2=(m+n)(m−n)\displaystyle m^2-n^2=(m+n)(m-n) ]

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    Add and subtract the two given relations.\[m+n=2\tan\theta,\qquad m-n=2\sin\theta \] \[m^2-n^2=(m+n)(m-n) \] \[m^2-n^2=(2\tan\theta)(2\sin\theta) \] \[m^2-n^2=4\sin\theta\tan\theta \]Answer: \(\displaystyle m^2-n^2=4\sin\theta\tan\theta\)