The function does not exist for one-toone but not onto.
Since f = A\(\rightarrow\)  B, f is one-one \(\Rightarrow\) f must be onto   [\(\therefore\) n(A) = n(B)]

Given A = {1, 2, 3, 4}, and B = {a, b, c, d}
Let f = A \(\rightarrow\) B.
The function does not exist for not one-one but onto. Since f = A\(\rightarrow\)B, f is onto \(\Rightarrow\) f must be one one since n(A) = n(B)

Let f = {(1, b) (2, b) (3, c) (4, e)}
Different elements in A does not have different images in B
∴ f is not one- one
Now, Co-domain = {a, b, e, d}, Range = {b, e}
Co-domain ≠ range
∴ f is not onto. Hence f is neither one - one and nor onto.

Let us prove that \(\frac { { a }^{ 2 }-{ b }^{ 2 }+{ c }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } } =\frac { tanC }{ tanB } -tanC.tanB\)
LHS = \(\frac { { a }^{ 2 }-{ b }^{ 2 }+{ c }^{ 2 } }{ { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } } =\frac { { k }^{ 2 }{ sin }^{ 2 }A-{ k }^{ 2 }{ sin }^{ 2 }B+{ k }^{ 2 }{ sin }^{ 2 }C }{ { k }^{ 2 }{ sin }^{ 2 }A+{ k }^{ 2 }{ sin }^{ 2 }B-{ k }^{ 2 }{ sin }^{ 2 }C } \)
\(\frac { { sin }^{ 2 }A-{ sin }^{ 2 }B+{ sin }^{ 2 }C }{ { sin }^{ 2 }A-{ sin }^{ 2 }C+{ sin }^{ 2 }B } \)
= \(\frac { sin\left( A+B \right) sin\left( A-B \right) +{ sin }^{ 2 }C }{ sin\left( A+C \right) sin\left( A-C \right) +{ sin }^{ 2 }B } \)
= \(\frac { sinC\left[ sin\left( A-B \right) +sinC \right] }{ sinB\left[ sin\left( A-C \right) +sinB \right] } \)
= \(\frac { sinC\left[ sin\left( A-B \right) +sin\left( A+B \right) \right] }{ sinB\left[ sin\left( A-C \right) +sin\left( A+C \right) \right] } \)
= \(\frac { sinC.2sinAcosB }{ sinB.2sinAcosC } \)
= \(\frac { sinC }{ cosC } .\frac { cosB }{ sinB } \)
= tan C = RHS
Hence proved.

\(LHS=1+cos2x+cos4x+cos6x\quad \left[ \because 1+cos2x=2{ cos }^{ 2 }x \right] \)
\(=2{ cos }^{ 2 }x+cos4x+cos6x\)
\(=2{ cos }^{ 2 }x+2cos\left( \frac { 4x+6x }{ 2 } \right) .cos\left( \frac { 4x-6x }{ 2 } \right) \quad \left[ \because cosA+cosB=2cos\left( \frac { A+B }{ 2 } \right) .cos\left( \frac { A-B }{ 2 } \right) \right] \)
= 2cos²x + 2cos 5x.cos(-x)
= 2cos²x + 2cos 5x cos x
\(=2cosx\left( 2cos\frac { x+5x }{ 2 } .cos\left( \frac { x-5x }{ 2 } \right) \right) \)
= 2cos x.2cos 3x.cos(-2x)
= 4cos x cos 3x.cos 2x
= RHS   \(\quad \left[ \because cos(-\theta =cos\theta \right] \)

\(LHS=\frac { sin4x+sin2x }{ cos4x+cos2x } \)
\(=\frac { 2sin\left( \frac { 4x+2x }{ 2 } \right) .cos\left( \frac { 4x-2x }{ 2 } \right) }{ 2cos\left( \frac { 4x+2x }{ 2 } \right) .cos\left( \frac { 4x-2x }{ 2 } \right) } =\frac { sin3x.cosx }{ cos3x.cosx } =tan3x\) = RHS

Let ABC be a triangle with constant perimeter 2s. Thus s is constant.
We know that \(\triangle =\sqrt{s(s-a)(s-b)(s-c)}\)
Observe  that \(\triangle\) is maximum, when (s - a) (s - b) (s - c) is maximum.
Now, (s-a)(s-b)(s-c)\(\le\)\(({(s-a)+(s-b)+(s-c)\over 3})^3\) = \({s^3\over 27}\) [G.M\(\le\) A.M]
Thus, we get (s - a) (s - b) (s - c) \(\le\) \({s^3\over 27}\)
Equality occurs when s - a = s - b = s - c. That is, when a = b = c maximum of (s - a) (s - b) (s - c) is \({s^3\over 27}\)
Thus, for a fixed perimeter 2s, the area of a triangle is maximum when a = b = c.
Hence, for a fixed perimeter, the equilateral triangle has the maximum area and the maximum area is given by \(\triangle =\sqrt {s(s)^3\over 27}={s^2\over3\sqrt{3}}sq.units\)

​​​​​​The law of sine: \(({a\over sin A})=({b\over sin B})=({c\over sin C})=2R\)
LHS \(={{a^2+b^2}\over{a^2+c^2}}={{(2R \sin A)}^{2}+{(2R \sin B)}^{2}\over{{(2 R \sin A)}^{2}+{(2R \sin C)}^{2}}}\)
\(={{\sin^2 A+\sin^2 B}\over{\sin^2 A+\sin^2 C}}={{1-\cos^2 A+\sin^2 B}\over{1-\cos^2 A+\sin^2 C}}\)
\(={{1-({\cos}^{2}A-{\sin}^{2}B)}\over{1-({\cos}^{2}A-{\sin}^{2}C)}}={1-\cos(A+B)\cos(A-B)\over1-\cos(A+C)\cos(A-C)}\)
\(={1+\cos(A-B)\cos C\over 1+\cos (A-C)\cos B}\)

i. cos A + cos B + cos C = 2 cos \(({A+B\over 2})\) cos \(({A-B\over 2})\) + cos C
= 2cos \(({\pi \over2}-{c\over2})cos ({A\over2}-{B\over2})+cos C \ (\frac{A+B+C}{2}=\frac{\pi}{2})\)
= 2sin \(({c\over2})cos ({A\over2}-{B\over2})+1-2 sin^2( {C\over 2})\)
= 1 + 2sin \(({c\over2})[cos ({A\over2}-{B\over2})- sin( {C\over 2})]\)
= 1 + 2sin \(({c\over2})[cos ({A\over2}-{B\over2})- cos({\pi\over2}- {C\over 2})]\)
= 1 + 2sin \(({c\over2})[cos ({A\over2}-{B\over2})- cos({A\over2}+ {B\over 2})]\)
= 1 + 4sin \(({A\over 2})sin({B\over2})sin({C\over 2})\)
ii. Let u = sin \(({A\over 2})sin({B\over2})sin({C\over 2})\)
\(=-{1\over2}[cos({A+B\over2})-cos({A-B\over2})]sin{c\over2}\)
\(=-{1\over2}[cos({A+B\over2})-cos({A-B\over2})]cos{A+B\over2}\)
\(=cos^2{A+B\over2}-cos{A-B\over2}cos{A+B\over2}+2u=0,\) which is quadratic in cos \({A+B\over2}\)
Since cos \({A+B\over2}\)  is real number, the above equation has a solution.
Thus, the discriminant b² - 4ac \(\ge\) 0, which gives
\(=cos^2{A-B\over2}-8u \ge 0 \Rightarrow u \le {1\over8}cos^2{A-B\over2}\le{1\over 8}\)
Hence, sin \(({A\over 2})sin({B\over2})sin({C\over 2})\le{1\over 8}\)
iii. From (i) and (ii), we have cos A + cos B + cos C > 1 and cos A + cos B + cos C ≤ 1 + 4 × \({1\over 8}\)
Thus, we get 1 < cos A + cos B + cos C \(\le\frac{3}{2}\)

The equation in distance form of the line passing through p(4, 1) and making an angle of 135^o with the positive x-axis is
\(\frac { x-4 }{ \cos { { 135 }^{ o } } } =\frac { y-1 }{ \sin { { 135 }^{ o } } } \)
Suppose it cuts 4x - y =0 at Q such that pQ = r.
Then, the co-ordinates of Q are
\(\frac { x-4 }{ \cos { { 135 }^{ o } } } =\frac { y-1 }{ \sin { { 135 }^{ o } } } =r\)
cos 135^o = cos (180^o - 45^o) = - cos 45^o = \(\frac { -1 }{ \sqrt { 2 } } \)
sin (135^o) = sin (180^o - 45^o) = - sin 45^o = \(\frac { -1 }{ \sqrt { 2 } } \)
Substituting these values in (1) we get,
\(\frac { x-4 }{ \frac { -1 }{ \sqrt { 2 } } } =\frac { y-1 }{ \frac { 1 }{ \sqrt { 2 } } } =r\)
\(\Rightarrow x=4-\frac { r }{ \sqrt { 2 } } ,y=1+\frac { r }{ \sqrt { 2 } } \)
\(\therefore\) Q is \(\left( 4-\frac { r }{ \sqrt { 2 } } ,1+\frac { r }{ \sqrt { 2 } } \right) \)
Clearly Q lies on 4x - y = 0
\(\therefore 16-\frac { 4r }{ \sqrt { 2 } } -1-\frac { r }{ \sqrt { 2 } } =0\)
\(\Rightarrow \frac { 5r }{ \sqrt { 2 } } =15\Rightarrow r=3\sqrt { 2 } \)
Hence, the required distance is \(3\sqrt2\)  units.

Let P(1, 2) and B(5, 3) are the given points.
By the property of reflector \(\angle \) XAB = \(\angle \) OAP = \(\theta\)
Let m₁ be the slope of the x-axis, m₂ and m₃ be the slopes of the lines AP and AB.
To find XAB,
Clearly m₁ = 0 Since it represents slope of x-axis.
\({ m }_{ 2 }=\frac { 2-0 }{ 1-x } =\frac { 2 }{ 1-x } and\left[ m=\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \right] \)
\({ m }_{ 3 }=\frac { 3-0 }{ 5-x } =\frac { 3 }{ 5-x } \)

\(tan\ \theta =\left| \frac { { m }_{ 1 }-{ m }_{ 3 } }{ 1+{ m }_{ 1 }.{ m }_{ 3 } } \right| \)
\(tan\ \theta =\left| \frac { 0-\frac { 3 }{ 5-x } }{ 1+0\left( \frac { 3 }{ 5 } -x \right) } \right| =\frac { 3 }{ 5-x } \)
To find \(\angle \)OAP,
\(tan(-\theta )=\left| \frac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| =\left| \frac { 0-\frac { 2 }{ 1-x } }{ 1+0\left( \frac { 2 }{ 1-x } \right) } \right| =\frac { 2 }{ 1-x } \)  [Since OAP is in the clockwise direction]
\(\Rightarrow \ -tan\theta =\frac { 2 }{ 1-x } \ \quad [\because tan\theta (-\theta )=tan\quad \theta ]\)
\(\Rightarrow \quad tan\quad \theta =\frac { -2 }{ 1-x } \)
From (1) and (2),
\(\frac { 3 }{ 5-x } =\frac { -2 }{ 1-x } \Rightarrow 3-3x=-10+2x\)
\(\Rightarrow\)  3 + 10 = 2x + 3x 
\(\Rightarrow\)  13 = 5x 
\(\Rightarrow \ x=\frac { 13 }{ 5 } \)
The required co-ordinates of A is \( \left( \frac { 13 }{ 5 } ,0 \right) \)

Given f(x) = \(\frac { \sqrt { 4-{ x }^{ 2 } } }{ \sqrt { { x }^{ 2 }-9 } } \)
When x = 2, f(x) = 0
When x = -2, f(x) = 0
For all the other values, we get negative value in the square root which is not possible.
\(\therefore\) Domain = {2, -2}

\(LHS=sin\frac { \theta }{ 2 } sin\frac { 7\theta }{ 2 } +sin\frac { 3\theta }{ 2 } sin\frac { 11\theta }{ 2 } \)
\(=\frac { 1 }{ 2 } \left[ cos\left( \frac { \theta }{ 2 } -\frac { 7\theta }{ 2 } \right) -cos\left( \frac { \theta }{ 2 } -\frac { 7\theta }{ 2 } \right) \right] +\frac { 1 }{ 2 } \left[ cos\left( \frac { 3\theta }{ 2 } -\frac { 11\theta }{ 2 } \right) -cos\left( \frac { 3\theta }{ 2 } -\frac { 11\theta }{ 2 } \right) \right] \)
\(=\frac { 1 }{ 2 } \left[ cos(-3\theta )-cos(4\theta )+cos(-4\theta )-cos7\theta \right] \)
\(=\frac { 1 }{ 2 } \left[ cos3\theta -cos4\theta +cos4\theta -cos7\theta \right] \quad \quad \left[ \because cos(-\theta )=cos\theta \right] \)
\(=\frac { 1 }{ 2 } \left[ cos3\theta -cos7\theta \right] =\frac { 1 }{ 2 } \left[ 2sin\left( \frac { 3\theta +7\theta }{ 2 } \right) .sin\left( \frac { 7\theta -3\theta }{ 2 } \right) \right] \)
\(=sin5\theta .sin2\theta =RHS.\)