SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Trigonometric Functions

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EXERCISE 3.3 1–10 (part 3 of 6)

  1. Prove that:

    Exercise 1

    sin⁡2π6+cos⁡2π3−tan⁡2π4=−12\displaystyle \sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}

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    Substitute the standard values. From the table of trigonometric ratios, \[\sin\frac{\pi}{6}=\frac{1}{2},\qquad \cos\frac{\pi}{3}=\frac{1}{2},\qquad \tan\frac{\pi}{4}=1.\]Taking the left-hand side, \[\text{LHS}=\sin^{2}\frac{\pi}{6}+\cos^{2}\frac{\pi}{3}-\tan^{2}\frac{\pi}{4}=\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}-(1)^{2}\] \[=\frac{1}{4}+\frac{1}{4}-1=\frac{1}{2}-1=-\frac{1}{2}=\text{RHS}.\]Hence \(\displaystyle \sin^{2}\dfrac{\pi}{6}+\cos^{2}\dfrac{\pi}{3}-\tan^{2}\dfrac{\pi}{4}=-\dfrac{1}{2}\), as required.

  2. Exercise 2

    2sin⁡2π6+cosec⁡27π6cos⁡2π3=32\displaystyle 2 \sin ^{2} \frac{\pi}{6}+\operatorname{cosec}^{2} \frac{7 \pi}{6} \cos ^{2} \frac{\pi}{3}=\frac{3}{2}

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    Standard values with quadrant reduction.Rule: an angle outside the first quadrant is written as \(\displaystyle n\cdot\frac{\pi}{2}\pm\theta\) with \(\displaystyle \theta\) acute; the value is the corresponding value of \(\displaystyle \theta\), with the sign fixed by the quadrant.\[\sin\frac{\pi}{6}=\frac{1}{2}\;\Rightarrow\;2\sin^{2}\frac{\pi}{6}=2\cdot\frac{1}{4}=\frac{1}{2}\]Now \(\displaystyle \frac{7\pi}{6}=\pi+\frac{\pi}{6}\) lies in the third quadrant, where sine is negative, so\[\sin\frac{7\pi}{6}=-\frac{1}{2}\;\Rightarrow\;\operatorname{cosec}\frac{7\pi}{6}=-2\;\Rightarrow\;\operatorname{cosec}^{2}\frac{7\pi}{6}=4\]and \(\displaystyle \cos\frac{\pi}{3}=\frac{1}{2}\Rightarrow\cos^{2}\frac{\pi}{3}=\frac{1}{4}\).Substituting,\[\text{LHS}=\frac{1}{2}+4\cdot\frac{1}{4}=\frac{1}{2}+1=\frac{3}{2}=\text{RHS}\]Hence \(\displaystyle 2\sin^{2}\dfrac{\pi}{6}+\operatorname{cosec}^{2}\dfrac{7\pi}{6}\cos^{2}\dfrac{\pi}{3}=\dfrac{3}{2}\).

  3. Exercise 3

    cot⁡2π6+cosec⁡5π6+3tan⁡2π6=6\displaystyle \cot ^{2} \frac{\pi}{6}+\operatorname{cosec} \frac{5 \pi}{6}+3 \tan ^{2} \frac{\pi}{6}=6

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    Standard values with quadrant reduction.\[\cot\frac{\pi}{6}=\sqrt{3}\;\Rightarrow\;\cot^{2}\frac{\pi}{6}=3\]Next, \(\displaystyle \frac{5\pi}{6}=\pi-\frac{\pi}{6}\) lies in the second quadrant, where sine is positive, so\[\sin\frac{5\pi}{6}=\sin\frac{\pi}{6}=\frac{1}{2}\;\Rightarrow\;\operatorname{cosec}\frac{5\pi}{6}=2\]and \(\displaystyle \tan\frac{\pi}{6}=\frac{1}{\sqrt{3}}\Rightarrow 3\tan^{2}\frac{\pi}{6}=3\cdot\frac{1}{3}=1\).Adding,\[\text{LHS}=3+2+1=6=\text{RHS}\]Hence \(\displaystyle \cot^{2}\dfrac{\pi}{6}+\operatorname{cosec}\dfrac{5\pi}{6}+3\tan^{2}\dfrac{\pi}{6}=6\).

  4. Exercise 4

    2sin⁡23π4+2cos⁡2π4+2sec⁡2π3=10\displaystyle 2 \sin ^{2} \frac{3 \pi}{4}+2 \cos ^{2} \frac{\pi}{4}+2 \sec ^{2} \frac{\pi}{3}=10

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    Standard values with quadrant reduction.\(\displaystyle \frac{3\pi}{4}=\pi-\frac{\pi}{4}\) lies in the second quadrant, where sine is positive, so \(\displaystyle \sin\frac{3\pi}{4}=\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}\). Then\[2\sin^{2}\frac{3\pi}{4}=2\cdot\frac{1}{2}=1,\qquad 2\cos^{2}\frac{\pi}{4}=2\cdot\frac{1}{2}=1\]Also \(\displaystyle \cos\frac{\pi}{3}=\frac{1}{2}\Rightarrow\sec\frac{\pi}{3}=2\), so\[2\sec^{2}\frac{\pi}{3}=2\cdot 4=8\]Adding,\[\text{LHS}=1+1+8=10=\text{RHS}\]Hence \(\displaystyle 2\sin^{2}\dfrac{3\pi}{4}+2\cos^{2}\dfrac{\pi}{4}+2\sec^{2}\dfrac{\pi}{3}=10\).

  5. Exercise 5

    Find the value of:
    (i)
    sin⁡75∘\displaystyle \sin 75^{\circ}
    (ii)
    tan⁡15∘\displaystyle \tan 15^{\circ}

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    NCERT’s answer
    (i)
    $\displaystyle \frac{\sqrt{3}+1}{2 \sqrt{2}}$ (ii) $\displaystyle 2-\sqrt{3}$
    Compound angle formulae.Split each angle into two angles whose values are known.(i) Write \(\displaystyle 75^{\circ}=45^{\circ}+30^{\circ}\) and use \(\displaystyle \sin(A+B)=\sin A\cos B+\cos A\sin B\):\[\sin 75^{\circ}=\sin 45^{\circ}\cos 30^{\circ}+\cos 45^{\circ}\sin 30^{\circ} =\frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\cdot\frac{1}{2}\]\[=\frac{\sqrt{3}+1}{2\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{4}\](ii) Write \(\displaystyle 15^{\circ}=45^{\circ}-30^{\circ}\) and use \(\displaystyle \tan(A-B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}\):\[\tan 15^{\circ}=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ}\tan 30^{\circ}} =\frac{1-\dfrac{1}{\sqrt{3}}}{1+\dfrac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\]Rationalising by multiplying numerator and denominator by \(\displaystyle \sqrt{3}-1\):\[\tan 15^{\circ}=\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3})^{2}-1^{2}}=\frac{4-2\sqrt{3}}{2}=2-\sqrt{3}\]\(\displaystyle \sin 75^{\circ}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\) and \(\displaystyle \tan 15^{\circ}=2-\sqrt{3}\).
  6. Prove the following:

    Exercise 6

    cos⁡(π4−x)cos⁡(π4−y)−sin⁡(π4−x)sin⁡(π4−y)=sin⁡(x+y)\displaystyle \cos \left(\frac{\pi}{4}-x\right) \cos \left(\frac{\pi}{4}-y\right)-\sin \left(\frac{\pi}{4}-x\right) \sin \left(\frac{\pi}{4}-y\right)=\sin (x+y)

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    Cosine of a sum.The left side has exactly the shape \(\displaystyle \cos A\cos B-\sin A\sin B=\cos(A+B)\), with \(\displaystyle A=\frac{\pi}{4}-x\) and \(\displaystyle B=\frac{\pi}{4}-y\).\[\text{LHS}=\cos\!\left[\left(\frac{\pi}{4}-x\right)+\left(\frac{\pi}{4}-y\right)\right] =\cos\!\left(\frac{\pi}{2}-(x+y)\right)\]Since \(\displaystyle \cos\!\left(\frac{\pi}{2}-\theta\right)=\sin\theta\),\[\text{LHS}=\sin(x+y)=\text{RHS}\]Hence \(\displaystyle \cos\!\left(\dfrac{\pi}{4}-x\right)\cos\!\left(\dfrac{\pi}{4}-y\right)-\sin\!\left(\dfrac{\pi}{4}-x\right)\sin\!\left(\dfrac{\pi}{4}-y\right)=\sin(x+y)\).

  7. Exercise 7

    tan⁡(π4+x)tan⁡(π4−x)=(1+tan⁡x1−tan⁡x)2\displaystyle \frac{\tan \left(\dfrac{\pi}{4}+x\right)}{\tan \left(\dfrac{\pi}{4}-x\right)}=\left(\frac{1+\tan x}{1-\tan x}\right)^{2}

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    Tangent of a sum and of a difference.Use \(\displaystyle \tan(A\pm B)=\dfrac{\tan A\pm\tan B}{1\mp\tan A\tan B}\) with \(\displaystyle A=\frac{\pi}{4}\), so \(\displaystyle \tan A=1\):\[\tan\!\left(\frac{\pi}{4}+x\right)=\frac{1+\tan x}{1-\tan x},\qquad \tan\!\left(\frac{\pi}{4}-x\right)=\frac{1-\tan x}{1+\tan x}\]Dividing the first by the second (that is, multiplying by the reciprocal),\[\text{LHS}=\frac{1+\tan x}{1-\tan x}\cdot\frac{1+\tan x}{1-\tan x} =\left(\frac{1+\tan x}{1-\tan x}\right)^{2}=\text{RHS}\]Hence \(\displaystyle \dfrac{\tan\!\left(\frac{\pi}{4}+x\right)}{\tan\!\left(\frac{\pi}{4}-x\right)}=\left(\dfrac{1+\tan x}{1-\tan x}\right)^{2}\).

  8. Exercise 8

    cos⁡(π+x)cos⁡(−x)sin⁡(π−x)cos⁡(π2+x)=cot⁡2x\displaystyle \frac{\cos (\pi+x) \cos (-x)}{\sin (\pi-x) \cos \left(\dfrac{\pi}{2}+x\right)}=\cot ^{2} x

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    Allied-angle reduction.Reduce each factor to a function of \(\displaystyle x\):\[\cos(\pi+x)=-\cos x,\qquad \cos(-x)=\cos x\] \[\sin(\pi-x)=\sin x,\qquad \cos\!\left(\frac{\pi}{2}+x\right)=-\sin x\]Substituting,\[\text{LHS}=\frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)}=\frac{-\cos^{2}x}{-\sin^{2}x}=\frac{\cos^{2}x}{\sin^{2}x}=\cot^{2}x=\text{RHS}\]Hence \(\displaystyle \dfrac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos\!\left(\frac{\pi}{2}+x\right)}=\cot^{2}x\).

  9. Exercise 9

    cos⁡(3π2+x)cos⁡(2π+x)[cot⁡(3π2−x)+cot⁡(2π+x)]=1\displaystyle \cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1

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    Allied-angle reduction, then a common denominator.Reduce each factor:\[\cos\!\left(\frac{3\pi}{2}+x\right)=\sin x,\qquad \cos(2\pi+x)=\cos x\] \[\cot\!\left(\frac{3\pi}{2}-x\right)=\tan x,\qquad \cot(2\pi+x)=\cot x\](For the third one, \(\displaystyle \cos\!\left(\frac{3\pi}{2}-x\right)=-\sin x\) and \(\displaystyle \sin\!\left(\frac{3\pi}{2}-x\right)=-\cos x\), so their ratio is \(\displaystyle \tan x\).)Substituting,\[\text{LHS}=\sin x\cos x\left(\tan x+\cot x\right) =\sin x\cos x\left(\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}\right)\]\[=\sin x\cos x\cdot\frac{\sin^{2}x+\cos^{2}x}{\sin x\cos x}=\sin^{2}x+\cos^{2}x=1=\text{RHS}\]Hence \(\displaystyle \cos\!\left(\dfrac{3\pi}{2}+x\right)\cos(2\pi+x)\left[\cot\!\left(\dfrac{3\pi}{2}-x\right)+\cot(2\pi+x)\right]=1\).

  10. Exercise 10

    sin⁡(n+1)xsin⁡(n+2)x+cos⁡(n+1)xcos⁡(n+2)x=cos⁡x\displaystyle \sin (n+1) x \sin (n+2) x+\cos (n+1) x \cos (n+2) x=\cos x

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    Cosine of a difference.The left side is \(\displaystyle \cos A\cos B+\sin A\sin B=\cos(A-B)\), with \(\displaystyle A=(n+2)x\) and \(\displaystyle B=(n+1)x\).\[\text{LHS}=\cos\big[(n+2)x-(n+1)x\big]=\cos x=\text{RHS}\]Hence \(\displaystyle \sin(n+1)x\,\sin(n+2)x+\cos(n+1)x\,\cos(n+2)x=\cos x\).