# Current Affairs JEE Main & Advanced

#### Integrals of the Form $\int{\frac{dx}{a+b\,cos\,x}}$ and $\int{\frac{dx}{a+b\,sin\,x}}$

To evaluate such form of integrals, proceed as follows:     (1) Put $\cos x=\frac{1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}\text{ and }\sin x=\frac{2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}.$     (2) Replace $1+{{\tan }^{2}}\frac{x}{2}$in the numerator by ${{\sec }^{2}}\frac{x}{2}.$     (3) Put $\tan \frac{x}{2}=t$ so that $\frac{1}{2}{{\sec }^{2}}\frac{x}{2}dx=dt.$     (4) Now, evaluate the integral obtained which will be of the form$\int{\frac{dt}{a{{t}^{2}}+bt+c}}$ by the method discussed earlier.     (i) $\int{\frac{dx}{a+b\cos x}}$     Case I : When $a>b$,  then     $\int{\frac{dx}{a+b\cos x}=\frac{2}{\sqrt{{{a}^{2}}-{{b}^{2}}}}\,\,{{\tan }^{-1}}\left( \sqrt{\frac{a-b}{a+b}}\tan \frac{x}{2} \right)}+c$     Case II : When $a<b$, then     $\int{\frac{dx}{a+b\cos x}}=\frac{1}{\sqrt{{{b}^{2}}-{{a}^{2}}}}\log \left| \frac{\sqrt{b-a}\tan \frac{x}{2}+\sqrt{b+a}}{\sqrt{b-a}\tan \frac{x}{2}-\sqrt{b+a}} \right|+c$     Case III : When $a=b$, then $\int{\frac{dx}{a+b\cos x}}=\frac{1}{a}\tan \frac{x}{2}+c$.     (ii) $\int{\frac{dx}{a+b\sin x}}$     Case I : When ${{a}^{2}}>{{b}^{2}}$ or $a>0$ and $a>b$,  then     $\int{\frac{dx}{a+b\sin x}=\frac{2}{\sqrt{{{a}^{2}}-{{b}^{2}}}}}{{\tan }^{-1}}\left[ \frac{a\tan \frac{x}{2}+b}{\sqrt{{{a}^{2}}-{{b}^{2}}}} \right]+c$     Case II : When ${{a}^{2}}<{{b}^{2}}$, then     $\int{\frac{dx}{a+b\sin x}=\frac{1}{\sqrt{{{b}^{2}}-{{a}^{2}}}}}\log \left| \frac{(a\tan \frac{x}{2}+b)-(\sqrt{{{b}^{2}}-{{a}^{2}}})}{(a\tan \frac{x}{2}+b)+\sqrt{{{b}^{2}}-{{a}^{2}}}} \right|+c$     Case III : When ${{a}^{2}}={{b}^{2}}$     In this case, either $b=a$ or $b=-a$     (a) When $b=a$, then      $\int{\frac{dx}{a+b\sin x}}=\frac{-1}{a}\cot \left( \frac{\pi }{4}+\frac{x}{2} \right)\,+c=\frac{1}{a}[\tan x-\sec x]+c$     (b) When $b=-a$, then $\int{\frac{dx}{a+b\sin x}=\frac{1}{a}\tan \left( \frac{\pi }{4}+\frac{x}{2} \right)}+c$.

#### Evaluation of the Various Forms of Integrals by Use of Standard Results

(1) $\int_{{}}^{{}}{\frac{dx}{a{{x}^{\mathbf{2}}}+bx+c}}$ :     Working rule : We write $\int_{{}}^{{}}{\frac{dx}{a{{x}^{2}}+bx+c}}$     $=\frac{1}{a}\int_{{}}^{{}}{\frac{dx}{{{x}^{2}}+\frac{b}{a}x+\frac{c}{a}}}$ $=\frac{1}{a}\int_{{}}^{{}}{\frac{dx}{{{\left( x+\frac{b}{2a} \right)}^{2}}+\frac{c}{a}-\frac{{{b}^{2}}}{4c}}}$, which is of the form $\int_{{}}^{{}}{\frac{dx}{{{X}^{2}}-{{A}^{2}}},\,\int_{{}}^{{}}{\frac{dx}{{{X}^{2}}+{{A}^{2}}}}}\text{ or }\int_{{}}^{{}}{\frac{dx}{{{A}^{2}}-{{X}^{2}}}}$.     (2) $\int_{{}}^{{}}{\frac{dx}{\sqrt{a{{x}^{\mathbf{2}}}+bx+c}}}$ : This can be reduced to one of the forms of $\int_{{}}^{{}}{\frac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}$, $\int_{{}}^{{}}{\frac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}}$ or $\int_{{}}^{{}}{\frac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}}$.     (3) $\int_{{}}^{{}}{\sqrt{a{{x}^{\mathbf{2}}}+bx+c}\mathbf{ }}dx$ : This can be reduced to one of the forms of $\int_{{}}^{{}}{\sqrt{{{a}^{2}}-{{x}^{2}}}}\,dx,\,\int_{{}}^{{}}{\sqrt{{{x}^{2}}-{{a}^{2}}}\,dx\text{ or }\int_{{}}^{{}}{\sqrt{{{a}^{2}}+{{x}^{2}}}}\,dx}$.     (4) $\int_{{}}^{{}}{\frac{(px+q)\,dx}{a{{x}^{2}}+bx+c}},$ $\int_{{}}^{{}}{\frac{(px+q)\,dx}{\sqrt{a{{x}^{2}}+bx+c}}}$, $\int_{{}}^{{}}{(px+q)\,\sqrt{a{{x}^{2}}+bx+c}}\,dx$ :     For the evaluation of any of these integrals,     Put $px+q=A$ {differentiation of $(a{{x}^{2}}+bx+c)\}+B$     Find A and B by comparing the coefficients of like powers of x on the two sides.     In this way the integral breaks up into two parts.     (5) Integrals of the form  $\int_{{}}^{{}}{\frac{{{x}^{\mathbf{2}}}+\mathbf{1}}{{{x}^{\mathbf{4}}}+\mathbf{k}{{x}^{\mathbf{2}}}+\mathbf{1}}}$dx,   $\int_{{}}^{{}}{\frac{{{x}^{\mathbf{2}}}-\mathbf{1}}{{{x}^{\mathbf{4}}}+k{{x}^{\mathbf{2}}}+\mathbf{1}}}dx$,$\int_{{}}^{{}}{\frac{dx}{{{x}^{\mathbf{4}}}+k{{x}^{\mathbf{2}}}+\mathbf{1}}}$, where $k\in R$     Working Method   (i) To evaluate these types of integrals divide the numerator and denominator by ${{x}^{2}}$.     (ii) Put $x+\frac{1}{x}=t$ or $x-\frac{1}{x}=t$ as required.     (6) $\int_{{}}^{{}}{\frac{{{x}^{2}}+{{a}^{2}}}{{{x}^{4}}+k{{x}^{2}}+{{a}^{4}}}\,dx,\,\int_{{}}^{{}}{\frac{{{x}^{2}}-{{a}^{2}}}{{{x}^{4}}+k{{x}^{2}}+{{a}^{4}}}}\,dx}$, where $k$ is a constant, negative or zero.     These integrals can be obtained by dividing numerator and denominator by ${{x}^{2}}$, then putting $x-\frac{{{a}^{2}}}{x}=t$ and $x+\frac{{{a}^{2}}}{x}=t$ respectively.     (7) Substitution for some irrational functions     (i) $\int_{{}}^{{}}{\frac{dx}{\sqrt{(x-\alpha )\,(x-\beta )}},\,\int_{{}}^{{}}{\sqrt{\left( \frac{x-\alpha }{\beta -x} \right)}}\,dx}$     $\int_{{}}^{{}}{\sqrt{(x-\alpha )\,(\beta -x)}\,dx}$, Put $x=\alpha {{\cos }^{2}}\theta +\beta {{\sin }^{2}}\theta$.     (ii) $\int_{{}}^{{}}{\frac{dx}{(px+q)\,\sqrt{(ax+b)}}}$, Put $ax+b={{t}^{2}}$     (iii) $\int_{{}}^{{}}{\frac{dx}{(px+q)\,\sqrt{a{{x}^{2}}+bx+c}}}$, Put $px+q=\frac{1}{t}$     (iv) $\int_{{}}^{{}}{\frac{dx}{(p{{x}^{2}}+r)\,\sqrt{(a{{x}^{2}}+c)}}}$, at first $x=\frac{1}{t}$ and then $a+c{{t}^{2}}={{z}^{2}}$.     (8) Integrals of the form $\int{\frac{dx}{P\sqrt{Q}}}$, (where $\mathbf{P}$ and $\mathbf{Q}$ are linear or quadratic expressions in $\mathbf{x}$) : To evaluate such types of integrals, we have following substitutions according to the nature of expressions of $P$ and $Q$ in $x$ :   (i) When $Q$ is linear and $P$ is linear or quadratic, we put $Q={{t}^{2}}$.   (ii) When $P$ is linear and $Q$ is quadratic, we put $P=\frac{1}{t}.$     (iii) When both $P$ and $Q$ are quadratic, we put $x=\frac{1}{t}.$

#### Position of Centre of Gravity in Some Special Cases

(1) C.G. of a uniform rod : The C.G. of a uniform rod lies at its mid-point.     (2) C.G. of a uniform parallelogram : The C.G. of a uniform parallelogram is the point of inter-section of the diagonals.     (3) C.G. of a uniform triangular lamina : The C.G. of a triangle lies on a median at a distance from the base equal to one third of the medians.     (4) Some Important points to remember (i) The C.G. of a uniform tetrahedron lies on the line joining a vertex to the C.G. of the opposite face, dividing this line in the ratio 3 : 1.     (5) The C.G. of a right circular solid cone lies at a distance $\frac{h}{4}$ from the base on the axis and divides it in the ratio 3 : 1.     (6) The C.G. of the curved surface of a right circular hollow cone lies at a distance $\frac{h}{3}$ from the base on the axis and divides it in the ratio 2 : 1     (7) The C.G. of a hemispherical shell at a distance $a/2$ from the centre on the symmetrical radius.     (8) The C.G. of a solid hemisphere lies on the central radius at a distance $\frac{3a}{8}$ from the centre where a is the radius.     (9) The C.G. of a circular arc subtending an angle $2\alpha$ at the centre is at a distance $\frac{a\sin \alpha }{\alpha }$ from the centre on the symmetrical radius, a being the radius, and $\alpha$ in radians.     (10) The C.G. of a sector of a circle subtending an angle $2\alpha$ at the centre is at a distance $\frac{2a}{3}\frac{\sin \alpha }{\alpha }$from the centre on the symmetrical radius, a being the radius and $\alpha$  in radians.     (11) The C.G. of the semi circular arc lies on the central radius at a distance of $\frac{2a}{\pi }$ from the boundary diameter, where a is the radius of the arc.

#### Integration by Parts

(1) When integrand involves more than one type of functions : We may solve such integrals by a rule  which is known as integration by parts. We know that, $\frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx}$     $\Rightarrow \,\,d(uv)=udv+vdu$$\Rightarrow \,\,\int{d\,(uv)=\int{udv+\int{vdu}}}$     If $u$ and $v$ are two functions of $x,$ then $\int{\underset{I}{\mathop{u}}\,}\,\underset{II}{\mathop{v}}\,dx=u$$\int_{{}}^{{}}{v\,dx-}$ $\int{\{\frac{du}{dx}.\int{vdx\}dx}}$ i.e., the integral of the product of two functions = (First function) $\times$ (Integral of second function) – Integral of {(Differentiation of first function) $\times$ (Integral of second function)} Before applying this rule proper choice of first and second function is necessary. Normally we use the following methods :     (i) In the product of two functions, one of the function is not directly integrable (i.e.,$\log |x|,\,{{\sin }^{-1}}x,\,{{\cos }^{-1}}x,\,{{\tan }^{-1}}x$......etc), then we take it as the first function and the remaining function is taken as the second function.         (ii) If there is no other function, then unity is taken as the second function e.g. in the integration of $\int{{{\sin }^{-1}}x\,dx,\,\int{\log x\,dx,\,1}}$ is taken as the second function.     (iii) If both of the function are directly integrable then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable. Usually, we use the following preference order for the first function.     (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential). This rule is simply called as “ILATE”.     (2) Integral is of the form $\int{{{e}^{x}}\left\{ f\mathbf{(}x\mathbf{)}+f\,'(x) \right\}}\,dx$ : If the integral is of the form $\int{{{e}^{x}}\left\{ f(x)+f'(x) \right\}dx,}$ then by breaking this integral into two integrals integrate one integral by parts and keeping other integral as it is, by doing so, we get     (i) $\int{{{e}^{x}}\left[ f(x)+f'(x) \right]}\,dx={{e}^{x}}f(x)+c$     (ii) $\int{{{e}^{mx}}\left[ mf(x)+f'(x) \right]}\,dx={{e}^{mx}}f(x)+c$     (iii) $\int{{{e}^{mx}}\left[ f(x)+\frac{f'(x)}{m} \right]}\,dx=\frac{{{e}^{mx}}f(x)}{m}+c$     (3) Integral is of the form $\int_{{}}^{{}}{[x\,{f}'(x)+f(x)]\,dx}$ : If the integral is of the form $\int{[x\,f'(x)+f(x)]\,dx}$then by breaking this integral into two integrals, integrate one integral by parts and keeping other integral as it is, by doing so, we get,     $\int{[x\,f'(x)}+f(x)]\,dx=x\,f(x)+c$     (4) Integrals of the form $\int{{{e}^{ax}}sinbxdx,\,\int{{{e}^{ax}}}cosbx\,dx}$ :     $\int{{{e}^{ax}}\sin bx}=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}(a\sin bx-b\cos bx)+c$ $=\frac{{{e}^{ax}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\sin (bx-{{\tan }^{-1}}\frac{b}{a})+c$     $\int{{{e}^{ax}}.\,\cos bx\,dx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}(a\cos bx}+b\sin bx)+c$ $=\frac{{{e}^{ax}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\cos \,\left( bx-{{\tan }^{-1}}\frac{b}{a} \right)+c$     $\int{{{e}^{ax}}.\,\sin (bx+c)\,dx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left[ a\sin (bx+c)-b\cos (bx+c) \right]}+k$$=\frac{{{e}^{ax}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\sin \left[ (bx+c)-{{\tan }^{-1}}\left( \frac{b}{a} \right) \right]+k$     $\int{{{e}^{ax}}.\,\cos (bx+c)\,dx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left[ a\cos (bx+c)-b\sin (bx+c) \right]}+k$$=\frac{{{e}^{ax}}}{\sqrt{{{a}^{2}}+{{b}^{2}}}}\cos \left[ (bx+c)-{{\tan }^{-1}}\left( \frac{b}{a} \right) \right]+k$

#### Centre of Gravity of a Compound Body and Remainder

(1) Centre of Gravity of a compound body :  Let ${{G}_{1}}$,  ${{G}_{2}}$ be the centres of gravity of the two parts of a body and let ${{w}_{1}},{{w}_{2}}$ be their weights.     Let G be the centre of gravity of the whole body. Then, at G, acts the whole weight $({{w}_{1}}+{{w}_{2}})$of the body.     Join ${{G}_{1}}{{G}_{2}}$; then G must lie on ${{G}_{1}}$${{G}_{2}}$. Let O be any fixed point on ${{G}_{1}}{{G}_{2}}$. Let $O{{G}_{1}}={{x}_{1}},O{{G}_{2}}={{x}_{2}}$ and $OG=\bar{x}$. Taking moments about O, we have $({{w}_{1}}+{{w}_{2}})\bar{x}={{w}_{1}}{{x}_{1}}+{{w}_{2}}{{x}_{2}}$     or  $\bar{x}=\frac{{{w}_{1}}{{x}_{1}}+{{w}_{2}}{{x}_{2}}}{{{w}_{1}}+{{w}_{2}}}$.     (2) Centre of gravity of the remainder : Let w be the weight of the whole body. Let a part B of the body of weight ${{w}_{1}}$be removed so that a part A of weight $w-{{w}_{1}}$ is left behind.       Let G be the centre of gravity of whole body and ${{G}_{1}}$, the C.G. of portion B which is removed. Let ${{G}_{2}}$ be the C.G. of the remaining portion A. Let O be a point on ${{G}_{1}}{{G}_{2}}$ and let it be regarded as origin. Let $O{{G}_{1}}={{x}_{1}},OG=x$, $O{{G}_{2}}={{x}_{2}}$ Taking moments about O,     $(w-{{w}_{1}}){{x}_{2}}+{{w}_{1}}{{x}_{1}}=wx$     or ${{x}_{2}}=\frac{wx-{{w}_{1}}{{x}_{1}}}{w-{{w}_{1}}}$.

#### Centre of Gravity

The centre of gravity of a body or a system of particles rigidly connected together, is that point through which the line of action of the weight of the body always passes in whatever position the body is placed and this point is called centroid. A body can have one and only one centre of gravity.     ${{w}_{1}},{{w}_{2}},...........,\,{{w}_{n}}$ are the weights of the particles placed at the points ${{A}_{1}}({{x}_{1}},{{y}_{1}}),{{A}_{2}}({{x}_{2}},{{y}_{2}}),..........,{{A}_{n}}({{x}_{n}},{{y}_{n)}}$ respectively, then the centre of gravity $G(\bar{x},\bar{y})$ is given by $\bar{x}=\frac{\sum{{{w}_{1}}{{x}_{1}}}}{\sum{{{w}_{1}}}},\bar{y}=\frac{\sum{{{w}_{1}}{{y}_{1}}}}{\sum{{{w}_{1}}}}$.

#### Limiting Equilibrium on an Inclined Plane

Let a body of weight W be on the point of sliding down a plane which is inclined at an angle $\alpha$  to the horizon. Let R be the normal reaction and $\mu R$ be the limiting friction acting up the plane.     Thus, the body is in limiting equilibrium under the action of three forces : $R,\,\,\mu R$ and W.     Resolving the forces along and perpendicular to the plane, we have $\mu R=W\sin \alpha \text{ and }R=W\cos \alpha$     $\Rightarrow \frac{\mu R}{R}=\frac{W\sin \alpha }{W\cos \alpha }\Rightarrow \mu =\tan \alpha$ $\Rightarrow \tan \lambda =\tan \alpha \Rightarrow \alpha =\lambda$     Thus, if a body be on the point of sliding down an inclined plane under its own weight, the inclination of the plane is equal to the angle of the friction.     (1) Least force required to pull a body up an inclined rough plane : Let a body of weight W be at  point $A,\,\,\alpha$ be the inclination of rough inclined plane to the horizontal and $\lambda$ be the angle of friction. Let P be the force acting at an angle $\theta$ with the plane required just to move body up the plane.     $P=W\frac{\sin (\alpha +\lambda )}{\cos (\theta -\lambda )}$,       $\left\{ \because \mu =\tan \lambda \right\}$     Clearly, the force P is least when $\cos (\theta -\lambda )$is maximum, i.e. when $\cos (\theta -\lambda )=1$, i.e. $\theta -\lambda =0$or $\theta =\lambda$. The least value of P is $W\sin (\alpha +\lambda )$     (2) Least force required to pull a body down an inclined plane : Let a body of weight W be at the point A, a be the inclination of rough inclined plane to the horizontal and l be the angle of friction. Let P be the force acting an angle q with the plane, required just to move the body up the plane.     $P=\frac{W\sin (\lambda -\alpha )}{\cos (\theta -\lambda )}$,        $[\because \mu =\tan \lambda ]$     Clearly, P is least when $\cos (\theta -\lambda )$ is maximum, i.e. when $\theta -\lambda =0$ or $\theta =\lambda$. The least value of P is W$\sin (\lambda -\alpha )$.
• If $\alpha =\lambda$, then the body is in limiting equilibrium and is just on the point of moving downwards.

• If $\alpha <\lambda$, then the least force required to move the body down the plane is $W\sin (\lambda -\alpha )$.

• If $\alpha =\lambda ,\,\alpha >\lambda \text{ or }\alpha <\lambda$, then the least force required to move the body up the plane is $W\sin (\alpha +\lambda )$.

• If $\alpha >\lambda ,$then the body will move down the plane under the action of more...

#### Coefficient of Friction

When one body is in limiting equilibrium in contact with another body, the constant ratio which the limiting force of friction bears to normal reaction at their point of contact, is called the coefficient of friction and it is generally denoted by $\mu$.     Thus, $\mu$ is the ratio of the limiting friction and normal reaction.     Hence, $\mu =\tan \lambda =\frac{\text{Maximum force of friction}}{\text{Normal reaction}}$     $\Rightarrow \mu =\frac{F}{R}\Rightarrow F=\mu R,$ where F is the limiting friction and R is the normal reaction.
• The value of $\mu$ depends on the substance of which the bodies are made and so it differs from one body to the other. Also, the value of $\mu$ always lies between 0 and 1. Its value is zero for a perfectly smooth body.

• Cone of friction : A cone whose vertex is at the point of contact of two rough bodies and whose axis lies along the common normal and whose semi-vertical angle is equal to the angle of friction is called cone of friction.

#### Friction

Friction is a retarding force which prevent one body from sliding on another. It is, therefore a reaction.     When two bodies are in contact with each other, then the property of roughness of the bodies by virtue of which a force is exerted between them to resist the motion of one body upon the other is called friction and the force exerted is called force of friction.     (1) Friction is a self adjusting force : Let a horizontal force P pull a heavy body of weight W resting on a smooth horizontal table. It will be noticed that up to a certain value of P, the body does not move. The reaction R of the table and the weight W of the body do not have any effect on the horizontal pull as they are vertical. It is the force of friction F, acting in the horizontal direction, which balances P and prevents the body from moving.     As P is increased, F also increases so as to balance P. Thus F increases with P. A stage comes when P just begins to move the body. At this stage F reaches its maximum value and is equal to the value of P at that instant. After that, if P is increased further, F does not increase any more and body begins to move.     This shows that friction is self adjusting, i.e. amount of friction exerted is not constant, but increases gradually from zero to a certain maximum limit.     (2) Statical friction : When one body tends to slide over the surface of another body and is not on the verge of motion then the friction called into play is called statical friction.     (3) Limiting friction : When one body is on the verge of sliding over the surface of another body then the friction called into play is called limiting friction.     (4) Dynamical friction : When one body is actually sliding over the surface of another body the friction called into play is called dynamical friction. Dynamical friction is of two types i.e., sliding friction and Rolling friction.     (5) Laws of friction     (i) Friction acts in the direction opposite to that in which the body is about to move.     (ii) The magnitude of the Friction between two bodies bears a constant ratio depends only on the nature of the materials of which these bodies are made.     (iii) Friction is independent of the shape and the area of the surfaces in contact, so long as the normal reaction between them is same, if the normal reaction is constant.     (iv) Limiting friction fs is directly proportional to the normal reaction R, i.e. ${{f}_{s}}\,\,\propto \,\,R$     ${{f}_{s}}={{\mu }_{s}}.R;$${{\mu }_{s}}=\frac{{{f}_{s}}}{R}$, where  ${{\mu }_{s}}$  is a constant which is called coefficient of statical friction.     In case of dynamic friction, ${{\mu }_{k}}=\frac{{{f}_{k}}}{R}$, more...

#### Equilibrium of Coplanar Forces

(1) If three forces keep a body in equilibrium, they must be coplanar.     (2) If three forces acting in one plane upon a rigid body keep it in equilibrium, they must either meet in a point or be parallel.     (3) When more than three forces acting on a rigid body, keep it in equilibrium, then it is not necessary that they meet at a point. The system of forces will be in equilibrium if there is neither translatory motion nor rotatory motion.     i.e., $X=0,\,\,Y=0,\,\,G=0$ or $R=0,\,\,G=0$.     (4) A system of coplanar forces acting upon a rigid body will be in equilibrium if the algebraic sum of their resolved parts in any two mutually perpendicular directions vanish separately, and if the algebraic sum of their moments about any point in their plane is zero.     (5) A system of coplanar forces acting upon a rigid body will be in equilibrium if the algebraic sum of the moments of the forces about each of three non-collinear points is zero.     (6) Trigonometrical theorem : If P is any point on the base BC of $\Delta ABC$ such that $BP\,:\,CP=m:n$.     Then, (i) $(m+n)\cot \theta =m\cot \alpha -n\cot \beta$,     where $\angle BAP=\alpha ,\angle CAP=\beta$.     (ii) $(n+m)\cot \theta =n\cot B-m\cot C$.

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