Trigonometric Functions Test 38

Given,

\(\cos ^{2} 15^{\circ}-\cos ^{2} 75^{\circ}\)

\(=\cos ^{2} 15^{\circ}-\cos ^{2}\left(90^{\circ}-15^{\circ}\right)\)\(\quad\quad\) \((\because \cos (90^{\circ}-\theta)=\sin \theta)\)

\(=\cos ^{2} 15^{\circ}-\sin ^{2} 15^{\circ}\)

\(=\cos \left(2 \times 15^{\circ}\right) \quad\left(\because \cos ^{2} \theta-\sin ^{2} \theta=\cos 2 \theta\right)\)

\(=\cos 30^{\circ}\)

\(=\frac{\sqrt{3}}{2}\)

The given equation can be solved as:

\( \sin 2 x+\cos x=0 \)

\( \Rightarrow 2 \sin x \cos x+\cos x=0 \)

\( \Rightarrow \cos x(2 \sin x+1)=0 \)

\( \Rightarrow \cos x=0 \) (or) \((2 \sin x+1)=0 \)

i.e \(\cos x=0 \) (or)\( \sin x=-\frac{1 }{ 2} \)

\( \Rightarrow \cos x=\cos \frac{\pi }{ 2} \) OR \(\sin x=\sin (-\frac{\pi }{ 6}) \)

\( \Rightarrow x=2 n \pi+-\frac{\pi }{ 2} \) OR \( x=n \pi+(-1)^n \times(-\frac{\pi }{ 6})\)

Given, \(\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\)

\(\Rightarrow \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\sqrt{\left.\frac{(1+\cos \theta) \times(1+\cos \theta))}{(1-\cos \theta) \times(1+ \cos \theta) }\right)} \quad\) (By rationalization)

\(\Rightarrow \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\sqrt{\frac{(1+\cos \theta)^{2}}{1-\cos ^{2} \theta}}\)

\(\Rightarrow \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\sqrt{\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}}\)\(\quad \left(\because 1-\cos ^{2} x=\sin ^{2} x\right)\)

\(\Rightarrow \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\sqrt{\left(\frac{1+\cos \theta}{\sin  \theta}\right)^{2}}\)

\(\Rightarrow \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}=\frac{1+\cos \theta}{\sin  \theta}\)

\(\Rightarrow \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\) = \(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}\)

\(\Rightarrow \sqrt{\frac{1+\cos \theta}{1-\cos \theta}}\) = \(\operatorname{cosec} \theta+\cot \theta\) \(\quad\quad (\because\operatorname{cosec} x=\frac{1}{\sin x},\cot x=\frac{\cos x}{\sin x})\)

Given,

\(\operatorname{cosec} \theta-\cot \theta=\sqrt{3} \ldots\) (i)

We know the trigonometric identity, \(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta=1\)

\((\operatorname{cosec} \theta-\cot \theta)(\operatorname{cosec} \theta+\cot \theta)=1\)

Using equation (i) we get,

\(\Rightarrow \sqrt{3}(\operatorname{cosec} \theta+\cot \theta)=1\)

\(\Rightarrow \operatorname{cosec} \theta+\cot \theta=\frac{1}{\sqrt{3}}\)...(ii)

Adding (i) and (ii) we get,

\(\operatorname{cosec} \theta-\cot \theta+\operatorname{cosec} \theta+\cot \theta=\sqrt{3}+\frac{1}{\sqrt{3}} \)

\(\Rightarrow 2 \operatorname{cosec} \theta=\sqrt{3}+\frac{1}{\sqrt{3}}\)

\(\Rightarrow \operatorname{cosec} \theta=\frac{2}{\sqrt{3}}\)

\(\Rightarrow \theta=60^{\circ}\) or \(\theta=120^{\circ}\)

Substitute the value of \(\theta=60^{\circ}\) in equation (i),

\(\operatorname{cosec} 60^{\circ}-\cot 60^{\circ}=\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}}\)

\(=\frac{1}{\sqrt{3}} \neq \sqrt{3}\)

\(\Rightarrow \theta=60^{\circ}\) not satisfying the original equation.

Now, Substitute the value of \(\theta=120^{\circ}\) in equation (i),

\(\operatorname{cosec} 120^{\circ}-\cot 120^{\circ}=\frac{2}{\sqrt{3}}-\frac{-1}{\sqrt{3}}\)

\(=\frac{3}{\sqrt{3}}\)

\(=\sqrt{3}\)

\(\Rightarrow \theta=120^{\circ}\) satisfying the original equation.

Now,

\(\tan \theta+\sin \theta=\tan 120^{\circ}+\sin 120^{\circ}\)

\(=-\sqrt{3}+\frac{\sqrt{3}}{2}\)

\(=\frac{-2 \sqrt{3}+\sqrt{3}}{2}\)

\(=-\frac{\sqrt{3}}{2}\)

Given,

\(\frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}\)

We know that,

\(\operatorname{cosec}^{2} x=1+\cot ^{2} x\)

Therefore,

\(\frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}=\frac{\cot x+\operatorname{cosec} x-\operatorname{cosec}^{2} x-\cot ^{2} x}{\cot x-\operatorname{cosec} x+\operatorname{cosec}^{2} x-\cot ^{2} x}\)

\(\Rightarrow \frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}=\frac{\cot x+\operatorname{cosec} x-\operatorname{cosec}^{2} x-\cot ^{2} x}{\cot x-\operatorname{cosec} x+1}\)

\(\Rightarrow \frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}=\frac{\cot x+\operatorname{cosec} x-(\operatorname{cosec} x-\cot x)(\operatorname{cosec} x+\cot x)}{\cot x-\operatorname{cosec} x+1}\)

\(\Rightarrow \frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}=\frac{(\cot x+\operatorname{cosec} x)(\operatorname{cotx}-\operatorname{cosec} x+1)}{(\cot x-\operatorname{cosec} x+1)}\)

\(\Rightarrow \frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}=(\cot x+\operatorname{cosec} x)\)

\(\Rightarrow \frac{\cot x+\operatorname{cosec} x-1}{\cot x-\operatorname{cosec} x+1}=\frac{\cos x+1}{\sin x}\)

Given,

\(2 \sin ^{2} \mathrm{~A}+3 \cos ^{2} \mathrm{~A}=2\)

\(\Rightarrow 2 \sin ^{2} A+3\left(1-\sin ^{2} A\right)=2\)\(\quad\quad(\because\cos ^{2} A=1-\sin ^{2} A)\)

\(\Rightarrow 2 \sin ^{2} A+ 3-3\sin ^{2} A=2\)

\(\Rightarrow -\sin ^{2} A=-1\)

\(\Rightarrow \sin ^{2} A=1\)

\(\Rightarrow \sin A=1\)

\(\Rightarrow \sin A=\sin 90^{\circ}\)

\(\Rightarrow A=90^{\circ}\)

Now, we will find

\((\tan A-\cot A)^{2}=\left(\tan 90^{\circ}-\cot 90^{\circ}\right)^{2}\)

\(\Rightarrow(\tan A-\cot A)^{2}=(\infty-0)^{2}\)

\(\Rightarrow(\tan A-\cot A)^{2}=\infty\)

Given,

\(A+B=90^{\circ}\),

\( \Rightarrow B=90^{\circ}-A\)

Therefore,

\(\sqrt{\sin \mathrm{A} \sec \mathrm{B}-\sin \mathrm{A} \cos \mathrm{B}}\)

\(=\sqrt{\sin \mathrm{A} \sec (90^{\circ}-\mathrm{A})-\sin \mathrm{A} \cos (90^{\circ}-\mathrm{A})}\)\(\quad\quad(\because B=90^{\circ}-A)\)

\(=\sqrt{\sin \mathrm{A} \operatorname{cosec} \mathrm{A}-\sin \mathrm{A} \sin \mathrm{A}}\) \(\quad\quad(\because \sec \left(90^{\circ}-\theta\right)=\operatorname{cosec}\theta\) and \(\cos \left(90^{\circ}-\theta\right)=\sin \theta, \operatorname{cosec} \theta =\frac{1}{\sin \theta})\)

\(=\sqrt{1-\sin ^{2} \mathrm{~A}}\)

\(=\sqrt{\cos ^{2} \mathrm{~A}}\)\(\quad\quad(\because 1-\sin ^{2}\mathrm{A}=\cos ^{2}\mathrm{A} )\)

= \(\cos \mathrm{A}\)

Given,

\(\frac{\cos x-\sin x+1}{\cos x+\sin x-1}\)

Assuming this as equation (i)

\(\frac{\cos x-\sin x+1}{\cos x+\sin x-1}\)......(i)

Now rationalizing eqution (i)

\(\frac{\cos x-(\sin x-1)}{\cos x+(\sin x-1)} \times \frac{\cos x-(\sin x-1)}{\cos x-(\sin x-1)}\)

\(=\frac{[\cos x-(\sin x-1)]^{2}}{\cos ^{2} x-(\sin x-1)^{2}}\)

Now using identity \((a- b)^{2}=(a^{2} - 2 a b-b^{2})\) we get,

\(=\frac{\cos ^{2} x+(\sin x-1)^{2}-2 \cos x(\sin x-1)}{-\left(\sin ^{2} x-2 \sin x+1\right)}\)

\(=\frac{\cos ^{2} x+\sin ^{2} x-2 \sin x+1-2 \cos x(\sin x-1)}{1-\sin ^{2} x-\left(\sin ^{2} x-2 \sin x+1\right)}\)

\(\quad\quad(\because \cos ^{2} x+\sin ^{2} x\) = 1 and \(\cos ^{2} x\) = \(1-\sin ^{2} x)\)

\(=\frac{1-2 \sin x+1-2 \cos x(\sin x-1)}{1-\sin ^{2} x-\left(\sin ^{2} x-2 \sin x+1\right)}\)

\(=\frac{2-2 \sin x-2 \cos x(\sin x-1)}{1-\sin ^{2} x-\sin ^{2} x+2 \sin x-1}\)

\(=\frac{2(1-\sin x)+2 \cos x(1-\sin x)}{2 \sin x-2 \sin ^{2} x}\)

\(=\frac{(1-\sin x)(1+\cos x)}{\sin x(1-\sin x)}\)

\(=\frac{1+\cos x}{\sin x}\)

\(=\frac{1}{\sin x}+\frac{\cos x}{\sin x}\)\((\because \operatorname{cosec} x=\frac{1}{\sin x}, \cot x=\frac{\cos x}{\sin x})\)

\(=\operatorname{cosec} x+\operatorname{cot} x\)

Given,

\(\sin 2 \theta=\cos 3 \theta, 0<\theta<\pi / 2\)

As we know that, \(\sin 2 \theta=2 \sin \theta \cos \theta\) and \(\cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta\)

\(\Rightarrow 2 \sin \theta \cos \theta=4 \cos ^{3} \theta-3 \cos \theta \)

\(\Rightarrow 2 \sin \theta=4 \cos ^{2} \theta-3\)

\(\Rightarrow 2 \sin \theta=4\left(1-\sin ^{2} \theta\right)-3=4-4 \sin ^{2} \theta-3\)

\(\Rightarrow 4 \sin ^{2} \theta+2 \sin \theta-1=0\)

Comparing the above equation with quadratic equation \(a x^{2}+b x+c=0, a=4, b=2\) and \(c=-1\)

Now substituting the values in the quadratic formula \(\frac{x=\left(-b \pm \sqrt{b}^{2}-4 a c\right) }{2 a}\) we get,

\(\sin \theta=\frac{-2 \pm \sqrt{-2^{2}-4(4)(-1)}}{2(4)}\)

\(=\frac{-2 \pm \sqrt{4+16}}{8}\)

\(=\frac{-2 \pm \sqrt{20}}{8}\)

\(=\frac{-1 \pm \sqrt{5}}{4}\)

Thus, \(\sin \theta=\frac{-1 \pm \sqrt{5}}{4}\)

Since, \(0<\theta<\frac{\pi}{ 2} \Rightarrow \theta\) lies between \(0^{\circ}\) to \(90^{\circ} \Rightarrow\) all ratios are positive.

\(\Rightarrow \sin \theta=\frac{-1+\sqrt{5}}{4}\)

As we know that, \(\cos 2 \theta=\cos ^{2} \theta-\sin ^{2} \theta\)

\(=1-2 \sin ^{2} \theta\)

\(\Rightarrow \cos 2 \theta=1-2\left(\frac{-1+\sqrt{5}}{4}\right)^{2}\)

\(=\frac{1+\sqrt{5}}{4}\)

\(\therefore\) The the value of \(\cos 2 \theta \) is \(\frac{1+\sqrt{5}}{4}\)

Given,

\(\tan x=-\frac{1}{\sqrt{3}}\)

The general solution of the equation \(\tan x=\tan \alpha\) is given by:

\(x=n \pi+a\), where \(a \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) and \(n \in Z\)

As we know that,

\(\tan \frac{5 \pi}{6}=-\frac{1}{\sqrt{3}}\)

\(\Rightarrow \tan x=\tan \frac{5 \pi}{6}\)

As we know that, if \(\tan x=\tan \alpha\) then \(x=n \pi+a\), where \(a \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) and \(n \in Z\) \(\Rightarrow x=n \pi+(\frac{5 \pi}{6})\) where \(n \in Z\)

As we know that, if the solutions of a trigonometric equation for which \(0 \leq \mathrm{x}<2 \mathrm{~m}\) are called principal solution.

So, the principal solutions of the given equation are \(x=\frac{5 \pi}{6}\) and \(\frac{11 \pi}{6}\)