TRIGONOMETRIC RATIOS OF POSITIVE ACUTE ANGLES (PART-1)

TRIGONOMETRIC-RATIOS-OF-POSITIVE-ACUTE-ANGLES (PART-1)


$1.\,$ If $\,\,p\tan\theta=\tan p\theta,\,$ show that, $\,\,\frac{\sin^2p\theta}{\sin^2\theta}=\frac{p^2}{1+(p^2-1)\sin^2\theta}.$

Sol. $\,\,p\tan\theta=\tan p\theta \rightarrow(1) \\ \Rightarrow p\frac{\sin\theta}{\cos\theta}=\frac{\sin p\theta}{\cos p\theta} \\ \Rightarrow \frac{\sin^2p\theta}{\sin^2\theta}=p^2\frac{\cos^2p\theta}{\cos^2\theta} \\~~~~~~~~~~~~~~~=\frac{p^2}{\sec^2p\theta\cos^2\theta}\\~~~~~~~~~~~~~~~=\frac{p^2}{(1+\tan^2p\theta)\cos^2\theta}\\~~~~~~~~~~~~~~~=\frac{p^2}{(1+p^2\tan^2\theta)\cos^2\theta}\,\,[\text{By (1)}]\\~~~~~~~~~~~~~~~=\frac{p^2}{\cos^2\theta+p^2\sin^2\theta}\\~~~~~~~~~~~~~~~=\frac{p^2}{1-\sin^2\theta+p^2\sin^2\theta}\\ \Rightarrow \frac{\sin^2p\theta}{\sin^2\theta}=\frac{p^2}{1+(p^2-1)\sin^2\theta}$


$\,2(i)\,$ If $\,\,(1+\sin\alpha)(1+\sin\beta)(1+\sin\gamma)\\=(1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma),\,\,$ show that each side is equal to $\,\,\pm\cos\alpha\cos\beta\cos\gamma .$

Sol. Let $\,\,k=(1+\sin\alpha)(1+\sin\beta)(1+\sin\gamma)\\=(1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma) \\ \Rightarrow k^2=k.k \\ \Rightarrow k^2=\left[(1+\sin\alpha)(1+\sin\beta)(1+\sin\gamma)\right]\\  \times \left[(1-\sin\alpha)(1-\sin\beta)(1-\sin\gamma)\right] \\ \Rightarrow k^2=[(1+\sin\alpha)(1-\sin\alpha)] \\ \times[(1+\sin\beta)(1-\sin\beta)][(1+\sin\gamma)(1-\sin\gamma)] \\ \Rightarrow k^2=(1-\sin^2\alpha)(1-\sin^2\beta)(1-\sin^2\gamma) \\ \Rightarrow k^2=\cos^2\alpha.\cos^2\beta.\cos^2\gamma  \\ \Rightarrow k=\pm\cos\alpha\cos\beta\cos\gamma\,\,\text{(proved)}$

$\,2(ii)\,$ If $\,\,(\csc\alpha-1)(\csc\beta-1)(\csc\gamma-1)\\=(\csc\alpha+1)(\csc\beta+1)(\csc\gamma+1),\,\,$ show that each side is equal to $\,\,\pm\cot\alpha\cot\beta\cot\gamma .$

By $\,\,\csc \alpha\,\,$ we mean $\,\,\text{cosec}\,\alpha$. 

Sol. Let $\,\,k=(\csc\alpha-1)(\csc\beta-1)(\csc\gamma-1)\\=(\csc\alpha+1)(\csc\beta+1)(\csc\gamma+1)\\ \Rightarrow k^2=k.k \\ \Rightarrow k^2=\left[(\csc\alpha-1)(\csc\beta-1)(\csc\gamma-1)\right]\\ \times \left[(\csc\alpha+1)(\csc\beta+1)(\csc\gamma+1)\right] \\ \Rightarrow k^2=[(\csc\alpha-1)(\csc\alpha+1)] \\ \times[(\csc\beta-1)(\csc\beta+1)][(\csc\gamma-1)(\csc\gamma+1)] \\ \Rightarrow k^2=(1-\csc^2\alpha)(1-\csc^2\beta)(1-\csc^2\gamma) \\ \Rightarrow k^2=\cot^2\alpha.\cot^2\beta.\cot^2\gamma  \\ \Rightarrow k=\pm\cot\alpha\cot\beta\cot\gamma\,\,\text{(proved)}$

$\,3.\,$ Eliminate $\,\theta\,$ in each of the following :


$\,(i)\,\,l_1 \cos\theta+m_1\sin\theta+n_1=0,\\l_2\cos\theta+m_2\sin\theta+n_2=0.$

Sol.  $\,l_1 \cos\theta+m_1\sin\theta+n_1=0\rightarrow(1)\\l_2\cos\theta+m_2\sin\theta+n_2=0 \rightarrow(2)$

From $\,(1)\,$ and $\,(2)\,$, we get by cross-multiplication 

$\,\frac{\cos\theta}{m_1n_2-m_2n_1}=\frac{\sin\theta}{n_1l_2-n_2l_1}=\frac{1}{l_1m_2-l_2m_1} \\ \Rightarrow \cos\theta=\frac{m_1n_2-m_2n_1}{l_1m_2-l_2m_1} \\ \text{and}\,\,\,\sin\theta=\frac{n_1l_2-n_2l_1}{l_1m_2-l_2m_1}$

Now, we know $\,\,\sin^2\theta+\cos^2\theta=1 \\ \Rightarrow \left(\frac{n_1l_2-n_2l_1}{l_1m_2-l_2m_1}\right)^2+\left(\frac{m_1n_2-m_2n_1}{l_1m_2-l_2m_1}\right)^2=1 \\ \Rightarrow (n_1l_2-n_2l_1)^2+(m_1n_2-m_2n_1)^2\\=(l_1m_2-l_2m_1)^2$

$3(ii)\,\,a\sec\theta+b\tan\theta+c=0, \\ \,a'\sec\theta+b'\tan\theta+c'=0.$

Sol. $\quad a\sec\theta+b\tan\theta+c=0 \rightarrow(1) \\ a'\sec\theta+b'\tan\theta+c'=0\rightarrow(2)$

From $\,(1)\,$ and $\,(2)\,$, we get by cross multiplication, 

$\frac{\sec\theta}{bc'-b'c}=\frac{\tan\theta}{ca'-c'a}=\frac{1}{ab'-a'b}  \\ \Rightarrow \sec\theta=\frac{bc'-b'c}{ab'-a'b}\rightarrow(3) \\ \text{and}\,\,\tan\theta=\frac{ca'-c'a}{ab'-a'b}\rightarrow(4)$

We know, $\,\,\sec^2\theta-\tan^2\theta=1 \\ \Rightarrow \left(\frac{bc'-b'c}{ab'-a'b}\right)^2-\left(\frac{ca'-c'a}{ab'-a'b}\right)^2=1\,\,[\text{By (3),(4)}] \\ \Rightarrow \left(bc'-b'c\right)^2-\left(ca'-c'a\right)^2=\left(ab'-a'b\right)^2$

$3(iii)\,\, \cos^3\theta+3\cos\theta\sin^2\theta=x,\\\sin^3\theta+3\sin\theta\cos^2\theta=y.$

Sol. $\,\, \cos^3\theta+3\cos\theta\sin^2\theta=x\rightarrow(1)\\ \text{and}\,\, \sin^3\theta+3\sin\theta\cos^2\theta=y\rightarrow(2)$

Adding $\,(1)\,$ and $\,(2)\,$ we get, 

$x+y=\cos^3\theta+3\cos\theta\sin^2\theta+\sin^3\theta\\+3\sin\theta\cos^2\theta \\ \Rightarrow x+y=\cos^3\theta+3\cos^2\theta\sin\theta\\+3\cos\theta\sin^2\theta+\sin^3\theta \\ \Rightarrow x+y=(\cos\theta+\sin\theta)^3 \\ \Rightarrow (x+y)^{\frac 13}=(\cos\theta+\sin\theta) \\ \Rightarrow (x+y)^{\frac 23}=(\cos\theta+\sin\theta)^2 \rightarrow(3)$

Again, from $\,(1)\,$ and $\,(2)\,$ we get, 

$x-y=\cos^3\theta+3\cos\theta\sin^2\theta-\sin^3\theta\\-3\sin\theta\cos^2\theta \\ \Rightarrow x-y=\cos^3\theta-3\cos^2\theta\sin\theta\\+3\cos\theta\sin^2\theta-\sin^3\theta \\ \Rightarrow x-y=(\cos\theta-\sin\theta)^3 \\ \Rightarrow (x-y)^{\frac 13}=(\cos\theta-\sin\theta) \\ \Rightarrow (x-y)^{\frac 23}=(\cos\theta-\sin\theta)^2 \rightarrow(4)$

Adding $\,(3)\,$ and $\,(4)\,$ we get,

$\,\,(x+y)^{\frac 23}+(x-y)^{\frac 23}\\=(\cos\theta+\sin\theta)^2+(\cos\theta-\sin\theta)^2\\=2(\cos^2\theta+\sin^2\theta)\\=2$


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