Here, LHS   = sec⁡sec4θ- 1 sec⁡sec2θ-1           = 1cos4θ – 1 1cos2θ – 1           = 1-cos4θ 4θ × cos2θ 1-cos2θ   .....

Here,  LHS = cos8A + sin 8A          = (cos4A)2 + (sin 4A)2         = (cos4A - sin 4A)2+ 2cos4A.sin4A         =[ ( cos2A + sin2A) ( cos2A - sin2A)]2 + .....

LHS = cos6A - sin 6A        = (cos2A)3 - (sin2A)3        = ( cos2A - sin2A) [ ( cos2A)2 + cos2A.sin2A + (sin2A)2]        = cos2A [ (cos2A + sin2A)2 - 2cos2A.sin2A .....

Here, RHS = 3 - 4 cos2A + cos4A.          = 3 - 4(1 - 2sin2A) + 2cos22A - 1          = 2 - 4 + 8sin2A + 2(1 - 2sin2A)2           = - 2 + 8sin2A + 2( 1 - .....

LHS=     3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6 x + cos6 x)

= 3 [(sin x – cos x)2]2 + 6(sin2x + cos2x + 2 sin x cos x) + 4 [(sin2x)3 + (cos2x)3]

= 3 [sin2x + .....

LHS =sin3A / sinA - Cos3A / cosA = [ sin3A cosA - cos3A sinA]/(sinA cosA) = sin(3A - A)/(sinA cosA) = sin2A/(sinA cosA) = 2 sinA cosA/(sinA cosA) = 2  = .....

Here, We know,            22°+ 23° = 45 Taking cot on both the sides;             cot(22°+ 23°) = cot 45° or,       cot⁡cot 22°+cot⁡cot .....

Here,       50° + 20° = 70°. Taking tan on both the sides, then       tan (50° + 20°) = tan 70° or,  tan50° +tan20° 1-tan50° tan20° = tan (90° - 20°)  .....

Here, Sin75o ? We know, Sin 75° = Sin (30° + 45°)              = Sin30° cos 45°+ cos30° Sin 45°              =1 2 ×12 + .....

Here, tan15o ? We know,     15° = 60° - 45° Now, Operating both sides by btan,     tan15° = tan (60° - 45° )                 =tan⁡tan 60° - tan⁡ .....

Here, We know, sin105° = sin (60° + 45°)              = sin 60° cos45°+ cos 60°. sin105° = 32×12 + 12 × 12              = .....

Here, LHS  = sin 15°          = sin (45°-30°)          = sin 45° cos 30° - cos 45°sin30o         =12× 32 - 1 2 × 1 2     .....

Here, Sin 75° = Sin (45°+30°)              = Sin 45° cos 30°+Sin30° cos 45°              = 12×32+1 2 × 12         .....

Here, We have given; tan A= 56 and tan B = 111 We know that: tan( A+B) = tan⁡tanA+tan⁡tanB1- tan⁡tanA.tan⁡tanB                                     .....

Here, We know,       20° + 25° = 45°. Taking tan on both the sides then,        tan (20° + 25°) = tan 45° or,  tan 20° + ⁡tan 25° 1-tan 20° tan 25° .....

Here Given equation of circle is;        x2 + y2 – 2x + 4y – 4 = 0…………… (i) or, x2 – 2x + y2 + 4y – 4 = 0  or, x2 – 2x + 1 + y2 + 4y + 4 = 4 + 5  or, (x .....

Here,  To prove the points A(0, 2) and B(3, -1)  lie on the circle, their distance from the centre O(3, 2) should be same. The distance from centre (AO) =  √{(x - x)2+ (y - y)2} = √{(3 - 0)2 + (2 - .....

The name of the locus is .....

Here,  Given equation of lines              x2 + 2kxy + 4y2 = 0 Comparing it with              ax2+ 2hxy + by2 = 0  than, a = 1 , 2h = 2k  and b = .....

Here, Given equation,           3x2 – 6xy + (k + 4)y2 = 0  Comparing it with ax2 + 2hxy + by2 = 0  Then, a = 3, 2h = -6 and b = k + 4  ∴ h = -3 . The condition of coincident:      .....

Here, Given equations of lines,          (3 – 6)x2  - 8xy + my2 = 0 . Comparing it with ax2 + 2hxy + by2 = 0  Then, a = m – 6 , 2h – 8 .....

Here, Given equation of lines;      2x2 + 3xy + y2 = 0. or,2x2 + 2xy + xy + y2 = 0 or,2x(x + y) + y(x + y) = 0 or,(2x + y)(x + y) = 0 Either, x + y = 0 ………… (i)       or, 2x + y = 0 .....

Here,  Given equation of lines; x2 – 2xy cosecθ + y2 = 0 or, x2 – 2.x. y cosec θ + (y cosecθ)2 – (y cosecθ)2 + y2 = 0  or, (x – y cosec θ)2 – y2 cosec2 θ + y2 = 0 or, .....

Here,  Given equation of lines 5x2 – 8xy + py2 = 0  Comparing it with ax2 + 2xhy + by2 = 0 then,  a = 5, 2h = -8  i.e. h = -4 and b = p . Since the lines are perpendicular to each other,  or, a + b = 0  or, 5 + p .....

Given lines are:        x = 3y and 3x = y  Þ x - 3y = 0 and 3x – y= 0 . Combining these lines .....

Here, Equation of pair of lines is,          kx2 – 12xy + 3y2 = 0 Comparing it with         ax2 + 2hxy + by2 = 0  then, a = k, 2h = -12 .....

Here, We have given equation,          16x2 – 24xy + 9y2 = 0 Comparing it with ax2 + 2hxy + by2 = 0  Then,  a = 16 ,     2h = -24 ,     h = - 12,     b = 9 . We .....

Here, Given equation of lines;      6x2 – xy – y2 = 0 or, 6x2- 3xy + 2xy – y2 = 0 or, 3x(2x – y) + y(2x – y) = 0 or, (2x – y)(3x + y) = 0 Either,      2x – y = 0 or, 3x + y .....

Here, Given equation of lines is            5x2 – 6xy – 5y2 = 0  Comparing it with ax2 + 2hxy + by2 = 0  Then, a = 5 and b = -5 . Now,   a + b = 5 – 5 = 0 . It shows that the .....

Here,  Given equation of lines;         2x2 + 8xy - (m+1)y2 = 0   Comparing it with ax2 + 2hxy + by2=0         then, a = 2 , 2h = 8 .....

Here, Given equation of lines are; x2 + 4xy + y2 = 0 Comparing it with ax2 + 2hxy + by2 = 0 then, a = 1, 2h = 4 =  h = 2 and b = 1 Let θ be angle between the lines then, tan θ =±2h2- aba+b or, tan .....

Here,  Given equation of lines; 3x2 + 8xy + my2 = 0  Comparing it with ax2 +2hxy + by = 0,  Then,  a = 3, 2h = 8 and b = m . Since the lines are perpendicular so, a + b = 0  or, 3 + m = 0  ∴    m = -3 .....

Here, Given equation is x2 - 2xy tan2a - y2 = 0. or, x2 - 2.x.y tan2a + (y tan2a)2 - (y tan2a)2 -  y2 = 0. or, (x - y tan2a)2 - y2(tan22a + 1) = 0 or, (x - y tan2a)2 - y2(sec22a) = .....

The required equation is ax2 + 2hxy + by2 + 2gx + 2fy + c = .....

The single equation is (x - 3)(x + 3) = 0                                  or, x2 - 9 = .....

The required equation is xy = .....

The required equation is ax2 + 2hxy + by2 = .....

Here,  Given equation of lines; x2 – 2xy cosecθ + y2 = 0 Comparing it with ax2 + 2hxy + by2 = 0 then, a = 1, h =- cosecθ and b = 1 Let α be angle between the lines then,      tan .....

Here, Given equations of pair of lines; x2 + xy – ky = 0 Comparing it with ax2 + 2hxy + by2 = 0,  Then, a = 1, 2h = 1 => h = 1/2 and b = - k Let θ be the angle between lines then,      tan θ = .....

Here,  Given equation of lines is x2 – 5xy + 4y2 = 0  or, x2 – 4xy – xy + 4y2 = 0 or, x(x – 4y) – y(x – 4y) = 0  or, (x – 4y) (x – y) = 0  Either, x – 4y = 0 .....

Here, Given equation of the line is 2x2 + kxy + 3y2 = 0 Angle between the lines is 45°, k =? Comparing given equation with ax2 + 2xhy + by2 = 0 then, a = 2, 2h = k, i.e. h = k2 and b = 3. Let θ be the angle between lines then,   .....

Here,  Given equation of lines       2x2 – 3xy – 5y2 = 0  or, 2x2 – 5xy + 2xy – 5y2 = 0  or, x(2x – 5y) + y(2x – 5y) = 0  or, (2x – 5y) (x + y) = 0  Either, 2x .....

Here, Slope of line passing through (3,-4) and (-2, k) m1 = y2-  y1x2-  x1 = k+4-2-3 =  k+4-5 Slope of line having equation 5x+y = -3 m2= coeff.  of x   coeff.  of  y = - .....

If two lines with slopes m1 & m2 are parallel to each other, the relation between m1 & m2 is  m1 = .....

Here,      Given equation of lines are,             x - 5y = 8 and             6x + 11y = 7, The matrix from of given equations are: 1-5611xy = 87 or, AX = .....

Here,      9x - 8y = 12 and 2x + 3y = 17. The matrix form of given equations are: 9-823xy = 1217 or, AX = B...............i) We have, A = 9-823= 27 + 16= 43 Adj (A) = 38-29 We know that, A-1 = 14338-29 We know that, from(i)   .....

Here, given equations are              2x + 3y = 3 and               x + 2y = 1. Writing in matrix form, 2312 xy = 31 or, AX = .....

Here, Given equation of lines are,         5x + 3y = 9 and          x + 4y = 13. The matrix from of given equations are: 5-932xy = 476 or, AX = B ..............................i) We have, A = 5374= .....

Here,         3x + 5y = 24 and 5x = 2y + 9 The matrix form of given equations are: 3   55-2xy = 249 or, AX = B...............i) We have, A = 355-2= -6 - 25 = -31 Adj (A) = -2-5-53 We know that, A-1 = 1A Adj(A) = .....

Here,      2ab4 -2 5 = 6 0 or, - 4+5a- 2b+20 = 6 0 Equating the corresponding elements then, 5a – 4 = 6 or, 5a = 10 ∴ a     = 2 and 2b + 20 = 0 or, - 2b  = - 20 ∴ b = 10. Thus, the value of a and b .....