Prove that:
Question: Prove that: $\cos ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{27}{11}$ Solution: To Prove: $\cos ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{27}{11}$ Formula Used: $\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left(x \times \sqrt{1-y^{2}}+y \times \sqrt{1-x^{2}}\right)$ Proof: $\mathrm{LHS}=\cos ^{-1} \frac{4}{5}+\sin ^{-1} \frac{3}{5} \ldots(1)$ Let $\cos \theta=\frac{4}{5}$ Therefore $\theta=\cos ^{-1} \frac{4}{5} \ldots$ (2) From the figure, $\sin \theta=\frac{3}...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x^{3}-1}{x^{3}+x} d x$ Solution: $I=\int \frac{x^{3}-1}{x^{3}+x} d x=\int 1-\frac{x+1}{x^{3}+x} d x$ $=\int 1 \mathrm{dx}-\int \frac{\mathrm{x}+1}{\mathrm{x}^{3}+\mathrm{x}} \mathrm{dx}$ $\frac{\mathrm{x}+1}{\mathrm{x}\left(\mathrm{x}^{2}+1\right)}=\frac{\mathrm{A}}{\mathrm{x}}+\frac{\mathrm{Bx}+\mathrm{C}}{\mathrm{x}^{2}+1}$ $x+1=A\left(x^{2}+1\right)+(B x+C)(x)$ Equating constants $A=1$ Equating coefficients of $x$ $1=\mathrm{C}$ Equating ...
Read More →Show that cos (2 tan-1 1/7) = sin (4 tan-1 1/3).
Question: Show that cos (2 tan-11/7) = sin (4 tan-11/3). Solution: Taking L.H.S, we have L.H.S. $=\cos \left(2 \tan ^{-1} \frac{1}{7}\right)$ $=\cos \left(\cos ^{-1} \frac{1-\left(\frac{1}{7}\right)^{2}}{1+\left(\frac{1}{7}\right)^{2}}\right) \quad\left(\because 2 \tan ^{-1} x=\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\right)$ $=\cos \left(\cos ^{-1} \frac{48 / 49}{50 / 49}\right)$ $=\cos \left(\cos ^{-1} \frac{24}{25}\right)=\frac{24}{25} \quad\left(\because \cos \left(\cos ^{-1} x\right)=x, x \in[-1,1...
Read More →If 2 tan-1(cos θ) = tan-1 (2 cosec θ),
Question: If 2 tan-1(cos ) = tan-1(2 cosec ), then show that = /4. Solution: Given, 2 tan-1(cos ) = tan-1(2 cosec ) So, $\tan ^{\prime}\left(\frac{2 \cos \theta}{1-\cos ^{2} \theta}\right)=\tan ^{\prime}(2 \operatorname{cosec} \theta) \quad\left(\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)\right)$ $\frac{2 \cos \theta}{\sin ^{2} \theta}=2 \operatorname{cosec} \theta$ $\frac{2 \cos \theta}{\sin ^{2} \theta}=\frac{2}{\sin \theta}$ $\frac{\cos \theta}{\sin \theta}=1 \quad \Rig...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{3 x+5}{x^{3}-x^{2}-x+1} d x$ Solution: $I=\int \frac{3 x+5}{x^{3}-x^{2}-x+1} d x=\int \frac{3 x+5}{(x-1)^{2}(x+1)}$ $\frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+1}$ $3 x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^{2}$ Put $x=1$ $-3+5=4 C$ $2=4 \mathrm{C}$ $C=\frac{1}{2}$ Put $x=0$ $5=-A+B+C$ $A=\frac{1}{2}$ $\int \frac{3 x+5}{(x-1)^{2}(x+1)} d x=\frac{1}{2} \int \frac{d x}{x-1}+4 \int \frac{d x}{(x-1)^{2}}+\frac{1}{2} \int \fr...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{3 x+5}{x^{3}-x^{2}-x+1} d x$ Solution: $I=\int \frac{3 x+5}{x^{3}-x^{2}-x+1} d x=\int \frac{3 x+5}{(x-1)^{2}(x+1)}$ $\frac{3 x+5}{(x-1)^{2}(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+1}$ $3 x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^{2}$ Put $x=1$ $-3+5=4 C$ $2=4 \mathrm{C}$ $C=\frac{1}{2}$ Put $x=0$ $5=-A+B+C$ $A=\frac{1}{2}$ $\int \frac{3 x+5}{(x-1)^{2}(x+1)} d x=\frac{1}{2} \int \frac{d x}{x-1}+4 \int \frac{d x}{(x-1)^{2}}+\frac{1}{2} \int \fr...
Read More →Prove that:
Question: Prove that: $\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{12}{13}=\sin ^{-1} \frac{56}{65}$ Solution: To Prove: $\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{12}{13}=\sin ^{-1} \frac{56}{65}$ Formula Used: $\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left(x \times \sqrt{1-y^{2}}+y \times \sqrt{1-x^{2}}\right)$ Proof: $\mathrm{LHS}=\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{12}{13} \ldots$ (1) Let $\cos \theta=\frac{3}{5}$ Therefore $\theta=\cos ^{-1} \frac{3}{5} \ldots$ (2) From the figure, $\sin \theta=\...
Read More →Find the value of the expression
Question: Find the value of the expression sin (2 tan-11/3) + cos (tan-122). Solution: Given expression, sin (2 tan-11/3) + cos (tan-122) $\sin \left(2 \tan ^{-1} \frac{1}{3}\right)=\sin \left(\sin ^{-1} \frac{2 \times \frac{1}{3}}{1+\left(\frac{1}{3}\right)^{2}}\right)\left(\because 2 \tan ^{-1} x=\sin ^{-1} \frac{2 x}{1+x^{2}}\right)$ $=\sin \left(\sin ^{-1} \frac{2 / 3}{10 / 9}\right)$ $=\sin \left(\sin ^{-1} \frac{3}{5}\right)=\frac{3}{5}$$\left(\because \sin \left(\sin ^{-1} x\right)=x, x \...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)} d x$ Solution: $I=\int \frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)} d x$ $\frac{x^{2}}{\left(x^{2}+1\right)\left(3 x^{2}+4\right)}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{3 x^{2}+4}$ $x^{2}=(A x+B)\left(3 x^{2}+4\right)+(C x+D)\left(x^{2}+1\right)$ $=(3 A+C) x^{3}+(3 B+D) x^{2}+(4 A+C) x+4 B+D$ Equating similar terms $3 A+C=0$ $3 B+D=1$ $4 A+C=0$ $4 B+D=0$ Solving we get, $A=0, B=-1, ...
Read More →Find the real solution of the equation
Question: Find the real solution of the equation $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2}$ Solution: Given equation, $\tan ^{-1} \sqrt{x(x+1)}+\sin ^{-1} \sqrt{x^{2}+x+1}=\frac{\pi}{2}$ $\tan ^{-1} \sqrt{x(x+1)}=\frac{\pi}{2}-\sin ^{-1} \sqrt{x^{2}+x+1}$ $=\cos ^{-1} \sqrt{x^{2}+x+1}$ $=\tan ^{-1} \frac{\sqrt{-x^{2}-x}}{\sqrt{x^{2}+x+1}}$(From figure) $\sqrt{x(x+1)}=\frac{\sqrt{-x^{2}-x}}{\sqrt{x^{2}+x+1}}$ $x^{2}+x=0$ $x=0,-1$ Hence, the real solutions of the given tr...
Read More →Prove that:
Question: Prove that: $\sin ^{-1} \frac{1}{\sqrt{5}}+\sin ^{-1} \frac{2}{\sqrt{5}}=\frac{\pi}{2}$ Solution: To Prove: $\sin ^{-1} \frac{1}{\sqrt{5}}+\sin ^{-1} \frac{2}{\sqrt{5}}=\frac{\pi}{2}$ Formula Used: $\sin ^{-1} x+\sin ^{-1} y=\sin ^{-1}\left(x \times \sqrt{1-y^{2}}+y \times \sqrt{1-x^{2}}\right)$ Proof: $\mathrm{LHS}=\sin ^{-1} \frac{1}{\sqrt{5}}+\sin ^{-1} \frac{2}{\sqrt{5}}$ $=\sin ^{-1}\left(\frac{1}{\sqrt{5}} \times \sqrt{1-\left(\frac{2}{\sqrt{5}}\right)^{2}}+\frac{2}{\sqrt{5}} \ti...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x$ Solution: $I=\int \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)} d x$ $\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac{A x+B}{x^{2}+1}+\frac{C x+D}{x^{2}+4}$ $1=(A x+B)\left(x^{2}+4\right)+(C x+D)\left(x^{2}+1\right)$ $=(A+C) x^{3}+(B+D) x^{2}+(4 A+C) x+4 B+D$ Equating similar terms $A+C=0$ $B+D=0$ $4 A+C=0$ $4 B+D=1$ We get, $\mathrm{A}=0 \mathrm{~B}=\frac{1}{3} \mathrm{C}=0 \ma...
Read More →Prove that:
Question: Prove that: $\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}$ Solution: To Prove: $\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}$ Formula Used: $\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left(x y-\sqrt{1-x^{2}} \times \sqrt{1-y^{2}}\right)$ Proof: $\mathrm{LHS}=\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}$ $=\cos ^{-1}\left(\frac{4}{5} \times \frac{12}{13}-\sqrt{1-\left(\frac{4}{5}\right)^{2}} \times \sqrt{1-\left(\frac{12}{13}\right)^{...
Read More →Prove the following
Question: Show that 2 tan-1(-3) = /2 + tan-1(-4/3) Solution: Taking L.H.S = 2 tan-1(-3) = -2 tan-13 (∵ tan-1(-x) = tan-1x, x R) $=-2\left[\frac{\pi}{2}-\cot ^{-1} 3\right] \quad\left(\because \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right)$ $=-2\left[\frac{\pi}{2}-\tan ^{-1} \frac{1}{3}\right] \quad\left(\because \tan ^{-1} x=\cot ^{-1} \frac{1}{x}, x0\right)$ $=-\pi+2 \tan ^{-1} \frac{1}{3}$ $=-\pi+\tan ^{-1} \frac{2 \cdot \frac{1}{3}}{1-\left(\frac{1}{3}\right)^{2}} \quad\left(\because 2 \tan ^...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{2 \mathrm{x}}{\mathrm{x}^{3}-1} \mathrm{dx}$ Solution: $I=\int \frac{2 x}{x^{3}-1} d x=\int \frac{2 x}{(x-1)\left(x^{2}+x+1\right)} d x$ $\frac{2 x}{(x-1)\left(x^{2}+x+1\right)}=\frac{A}{x-1}+\frac{B x+C}{x^{2}+x+1}$ $2 x=A\left(x^{2}+x+1\right)+(B x+C)(x-1)$ $=(A+B) x^{2}+(A-B+C) x+(A-C)$ Equating constants, $A-C=0$ Equating coefficients of $x$ $2=A-B+C$ Equating coefficients of $x^{2}$ $0=A+B$ On solving, We get $A=\frac{2}{3} B=-\frac{2}{...
Read More →Find the value of
Question: Find the value of tan-1(tan 2/3). Solution: We know that, tan-1tan x = x, x (-/2, /2) $\tan ^{-1}\left(\tan \frac{2 \pi}{3}\right)=\tan ^{-1} \tan \left(\pi-\frac{\pi}{3}\right)=\tan ^{-1}\left(\tan \left(-\frac{\pi}{3}\right)\right)=-\frac{\pi}{3}$...
Read More →Prove that:
Question: Prove that: $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \tan ^{1} \frac{4}{3}$ Solution: To Prove: $\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}=\frac{1}{2} \tan ^{-1} \frac{4}{3} \Rightarrow 2\left(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}\right)=\tan ^{-1} \frac{4}{3}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y1$ Proof: $\mathrm{LHS}=2\left(\tan ^{-1} \frac{1}{4}+\tan ^{-1} \frac{2}{9}\right)$ $=2\left(\tan ^{-1}\...
Read More →Find the value of
Question: Find the value of tan-1(-1/3) + cot-1(1/3) + tan-1(sin (-/2)) Solution: Given, tan-1(-1/3) + cot-1(1/3) + tan-1(sin (-/2)) $=\tan ^{-1}\left(\tan \left(-\frac{\pi}{6}\right)\right)+\cot ^{-1}\left(\cot \frac{\pi}{3}\right)+\tan ^{-1}(-1)$ $=-\frac{\pi}{6}+\frac{\pi}{3}+\left(-\frac{\pi}{4}\right)=-\frac{\pi}{12}$...
Read More →Prove that:
Question: Prove that: $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$ Solution: To Prove: $\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y1$ Proof: $\mathrm{LHS}=\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{8}$ $=\tan ^{-1} \frac{1}{2}+\tan ^{-1}\left(\frac{\frac{1}{5}+\frac{1}{8}}{1-\left(\frac{1}{5} \times \f...
Read More →Prove that cot (π/4 – 2 cot-1 3) = 7
Question: Prove that cot (/4 2 cot-13) = 7 Solution: Re-writing the given, $\frac{\pi}{4}-2 \cot ^{-1} 3=\cot ^{-1} 7$ $2 \tan ^{-1} \frac{1}{3}=\frac{\pi}{4}-\tan ^{-1} \frac{1}{7}$ $2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}=\frac{\pi}{4}$ Now, $2 \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{7}$ $=\tan ^{-1} \frac{2 / 3}{1-(1 / 3)^{2}}+\tan ^{-1} \frac{1}{7}\left(\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right)$ $=\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{\...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{(x+1)^{2}\left(x^{2}+1\right)} d x$ Solution: $I=\frac{1}{(x+1)^{2}\left(x^{2}+1\right)}$ $\frac{1}{(x+1)^{2}\left(x^{2}+1\right)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C x+D}{x^{2}+1}$ $1=A(x+1)\left(x^{2}+1\right)+B\left(x^{2}+1\right)+(C x+D)(x+1)^{2}$ $=A x^{3}+A x^{2}+A x+A+B x^{2}+B+C x^{3}+2 C x^{2}+C x+D x^{2}+2 D+D$ $=(A+C) x^{3}+(A+B+2 C+D) x^{2}+(A+C+2 D) x+(A+B+D)$ Equating constants $1=A+B+D$ Equating coefficients of $x^{3}...
Read More →Prove that:
Question: Prove that: $\tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3=\pi$ Solution: To Prove: $\tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3=\pi$ Formula Used: $\tan ^{-1} x+\tan ^{-1} y=\pi+\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ where $x y1$ Proof: $\mathrm{LHS}=\tan ^{-1} 1+\tan ^{-1} 2+\tan ^{-1} 3$ $=\frac{\pi}{4}+\pi+\tan ^{-1}\left(\frac{2+3}{1-(2 \times 3)}\right)\{$ since $2 \times 3=61\}$ $=\frac{5 \pi}{4}+\tan ^{-1}\left(\frac{5}{-5}\right)$ $=\frac{5 \pi}{4}+\tan ^{-1}(-1)$ $=\frac{5 \pi}{4}-\fr...
Read More →Evaluate cos[cos-1(-√3/2) + π/6]
Question: Evaluate cos[cos-1(-3/2) + /6] Solution: $\cos \left[\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)+\frac{\pi}{6}\right]$ $=\cos \left[\cos ^{-1}\left(\cos \frac{5 \pi}{6}\right)+\frac{\pi}{6}\right]$ $\left(\because \cos \frac{5 \pi}{6}=\frac{-\sqrt{3}}{2}\right)$ $=\cos \left(\frac{5 \pi}{6}+\frac{\pi}{6}\right)$ $\left(\because \cos ^{-1} \cos x=x ; x \in[0, \pi]\right)$ $=\cos (\pi)=-1$...
Read More →Find the value of tan-1 [tan (5π/6)] + cos-1 [cos (13π/6)]
Question: Find the value of tan-1[tan (5/6)] + cos-1[cos (13/6)] Solution: We know that, tan-1tan x = x, x (-/2, /2) And, here tan-1tan (5/6) 5/6 as 5/6 (-/2, /2) Also, cos-1cos x = x; x [0, ] So, cos-1cos (13/6) 13/6 as 13/6 [0, ] Now, tan-1[tan (5/6)] + cos-1[cos (13/6)] = tan-1[tan ( /6)] + cos-1[cos (2 + /6)] = tan-1[ -tan /6] + cos-1[ -cos (7/6)] = tan-1[tan /6] + cos-1[cos (/6)] = /6 + /6 = 0...
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{1+x+x^{2}+x^{3}} d x$ Solution: $I=\int \frac{1}{1+x+x^{2}+x^{3}}=\int \frac{d x}{\left(x^{2}+1\right)(x+1)}$ $\frac{1}{\left(x^{2}+1\right)(x+1)}=\frac{A x+B}{x^{2}+1}+\frac{C}{x+1}$ $1=(A x+B)(x+1)+C\left(x^{2}+1\right)$ Equating constants $1=B+C$ Equating coefficients of $x$ $0=A+C$ Solving, we get $\mathrm{A}=-\frac{1}{2} \mathrm{~B}=\frac{1}{2} \mathrm{C}=\frac{1}{2}$ Thus $I=-\frac{1}{2} \int \frac{x d x}{x^{2}+1}+\frac{1}{2} \int \...
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