Here, Given A+B+C =180o. Now,  L.H.S. = sin A.cosB.cosC + sinB. cosA.cosC + sinC.cosA.cosB             =cosC( sinA cosB+ sinB cosA) + sinC cosA cosB             = cosC sin( A+B) + sinC .....

We have given,      A + B + C = πc  or, A + B =  πc - C or, cos(A + B) = cos(πc - C) or, cos(A + B) = - cosC or, cosA cosB - sinA sinB = - cosC or, cosA cosB + cosC = sinA sinB Squaring both .....

We have given,      A + B + C = 180o   or, A + B = 180o - C Operating both sides by tan, we get        tan(A + B) = tan( 180o - C) or,  tanA +  cotB  1 - tanA tanB = - tanC or,  .....

Here, we have given,     A + B + C = 180o ⟹ A + B = 180o - C  LHS = cos2A - cos2B - cos2C         =  2sin2A + 2B2 sin2B - 2A2 - cos2C        = 2sin(A + B) sin(B - A) - .....

We have given,      A + B + C = 180o   or, 2A + 2B = 180o - 2C Operating both sides by tan, we get        tan(2A + 2B) = tan( 180o - 2C) or,   tan2A+ tan2B 1 - tan2A tan2BcotA   .....

Here,  We have given,          A + B + C = 180o or,    A + B = 180 - C Now operating on both sides by cot we get          Cot(A + B) = cot(180 - C) or,    cotA cotB - 1cotA .....

Here,         P + Q + R =180° or, P + Q =180°- R Taking sin and cos on both the sides then,          sin (P + Q) = sin (180°- R) =   sin R         cos (P + .....

Here, LHS= cosec 2θ - cot 2θ        = 1⁡sin2θ - cos2θsin2θ        = 1-cos2θsin2θ        = 2 sin 2θ2sinθcos θ ⁡    .....

Here, LHS =  8 cos 10°. cos 50° cos 70°           = 4 cos 10°(2 cos 50°. cos 70°)           = 4 cos 10°[cos (50°+ 70°) + cos(50° - 70°)]     .....

Here, LHS = ⁡cos20°-sin20° cos20°+⁡sin20°          = cos20°-sin(90°-70°) cos20°+ sin(90°-70°)         = cos20°-⁡cos 70° cos20°+ ⁡cos 70° .....

Here, sin θ = 35 We know that, cos 2θ = 1- 2sin2 θ            = 1-2 (35)2            = 1- 2 × 925           = 725. Thus, the value of cos .....

Here,  LHS   = 2cos70°. cos20°                    = cos(70° + 20°) + cos(70° - 20°)            = cos90° + .....

Here, LHS = cosA-⁡cos5A sin 5A -⁡sinA        = 2sin(A+5A 2) sin(5A-A 2)2cos (5A+A 2)⁡sin(5A-A 2)        = sin3A Cos 3A        = tan 3A        = .....

Here, LHS = 4(cos320o + sin350o)         = 4cos320o + 4sin350o         = cos(3.20o) + 3cos20o + 3 sin50o - sin(3.50o)         =  cos60o + .....

Here,     cos15o .cos105o  =   12 (2cos15o .cos105 o) =   12 [cos(15o+105o) + cos(15o - 105o)] =   12 [ cos120o + cos90o] =   12 [ - 12  + .....

Here,    2sin15o .sin105 o = cos(15o - 105o) - cos(15o + 105o) =  cos(- 90o) - cos120o =  cos90o - cos120o =  0 -(- 12 ) =    12 , .....

Here,  LHS = 4 sinA sin(60o +A). sin(60o - A)          = 2sinA[2 sin(60o + A). sin(60o+ A) ]          = 2sinA[cos(60o + A - 60o + A) - cos(60o + A + 60o-A)]       .....

Here,  LHS = 4 cos(60o + A). sin(30o + A) . cosA          = 2[2 cos(60o + A). sin(30o+ A) ] cosA          = 2cosA[sin(60o + A + 30o + A) - sin(60o + A - 30o-A)]       .....

Here,  LHS = 4 cos(60o + A). cos(60o - A) . cosA          = 2[2 cos(60o + A). cos(60o - A) ] cosA          = 2cosA[cos(60 + A + 60 - A) + cos(60 + A - 60-A)]       .....

Here, L.H.S = sin⁡sin(θ 2) + sin⁡+ sinθ1+cosθ 2 + cos⁡cosθ = sin⁡sin(θ 2) + 2sin⁡+ 2sinθ 2cos⁡cosθ 21+cosθ 2 + 2cos 2⁡2cos 2θ 2 -1 = sin⁡sin θ 2 ( 1+2⁡( .....

Here, Sin36o - cos36o =Sin36o - cos(90 -54o) = Sin36o - sin54o   Using the formula of SinC - SinD, we get = 2Cos(36+54)/2.Sin(36-54)/2 = 2 cos45. sin(-9o) = 2 x 1/√2 (-sin9o) = .....

Here, LHS=  12 (cos 2θ - cos 8θ)        =  12 ×2 sin(2θ+8θ 2) sin(8θ-2θ 2)        =  sin 5θ. sin3θ       .....

Here, LHS = cos⁡cos40°-sin⁡sin30° sin⁡sin60°-cos⁡cos50° =   ⁡ cos⁡cos(90°-50°)-sin⁡sin30° ⁡ sin⁡sin(90°-30°)- cos⁡cos50° =  sin 50°⁡sin 50°- sin 30°cos .....

Here, LHS = sin 20° sin 30° sin 40° sin 80°         = sin 20° 12 in 40° sin 80°        = sin 20°× 14 × (2 sin 80° sin 40° )        = 14 .....

Here, LHS = tanθ + 2 tan 2θ + 4 cot 4 θ          = tan θ + 2 tan 2θ + 4 tan4θ          = tanθ+ 2 tan 2θ + 4 (1-tan 2⁡tan 22θ)2tan2θ .....

Here,   LHS = cos3  . cos 3θ + sin3 θ . sin 3θ. We know that,       cos 3θ = 4cos3 θ - 3 cos θ or, 4cos3 θ = cos3θ + 3 cos θ ∴ cos3θ = 1 4 ( .....

Here, LHS = ⁡ ⁡sinθ+sin2θ⁡ 1+⁡cosθ+cos2θ        = ⁡ sinθ+2⁡sinθ⁡cosθ 1+⁡cosθ+⁡ 2cos 2θ-1        = ⁡ sinθ1+2⁡cosθ ⁡ .....

Here, RHS = 1+ sinA -⁡cosA 1+ sinA+⁡cosA         =  (1- cosA)+Sin A (1+cosA)+Sin A         = 2⁡sin2 A 2+2⁡sin A 2 ⁡cos A 22⁡cos2 A 2+2sin A 2 cos A 2       .....

Here, RHS = 1 +sin ∝ -cos ∝ 1 +sin ∝ +cos ∝          = 1 -cos ∝ +sin ∝ 1 +cos ∝+sin ∝          = 2sin 2 ∝ 2  +2sin ∝ 2 cos ∝ 2 )2cos 2 .....

Here, Tan A2 = 34, SinA =? We know, Sin A = 2tan A21+tan 2 A 2         = 2×⁡× 341+3 4 2         = 341+916        = 322516       = 32×1625   .....

Here, cos ⁡ ∝ 3 = ⁡ ⁡ 1 2 So, sin ⁡ ∝ 3 = 1 –cos 2 ⁡ ∝ 3 = 1- ⁡ 1 2 2 =1- ⁡ 1 4 = ⁡ ⁡ 3 2 We know that, sin∝ = 3 sin ⁡ ∝ 3 - 4 sin3 ⁡ ∝ 3 = 3 × ⁡ ⁡ 32 - 4 ( ⁡ ⁡ 3 .....

Here, LHS =   1+ cosθ+sinθ 1⁡cosθ⁡+ sinθ  = ⁡ 2 cos 2 ⁡ θ 2+  2sin ⁡ θ 2 ⁡ cos ⁡ θ 2⁡ 2 sin 2 ⁡ θ 2+  2sin ⁡ θ 2  cos ⁡ θ 2 = .....

Here, sin A = 12 We know that, sin 3A = 3 sin A- 4 sin3 A            = 3 × 12 – 4( 12) 3            = 32 – 1 2            = 1. Thus, .....

Here,   LHS = cos3A.cos 3A + sin3 A.sin 3A          = 1 4 [4cos3A cos 3A + 4 sin3A sin 3A] ...............(i) We have, cos 3A = 4cos3 A - 3cos A .....

Here, LHS: = cosec θ - cot θ = ⁡ 1 sinθ - ⁡cosθ sinθ = ⁡ 1-⁡cosθ sinθ = 2 sin 2 ⁡ θ 2 2 sin ⁡ θ 2 ⁡cos ⁡ θ 2 = sin ⁡ θ 2 cos ⁡ θ 2 = .....

Here,   LHS = 8 (sin6 p + cos6 p)           = 8 [(cos2 p)3] + [(sin2 p)3 ]           = 8 [(cos2 p + sin2 p)3] - 3 cos2 p sin2 p (cos2 p +sin2 p )]          = 8 .....

Here, LHS = cosec 2θ - cot 2θ         = ⁡ 1 sin2θ - ⁡ ⁡cos2θ sin2θ         = ⁡ 1-cos2θ sin2θ         = 2 sin 2 θ 2 .....

Here,  LHS = {(1 + cos2θ)/sin2θ}. (1 + cosθ)/cosθ          = {(1 + cos2θ) (1 + cosθ)}/(sin2θcosθ)     We know,      1 + cos2θ = .....

We know,  cos2A = 1 - 2sin2A              = 1 - 2(1/2)2             = 1 - 1/2             = .....

Here, Replace B by A, we get sin2A = 2 sinA .....

Here, Tan2A = 2tanA / (1 - tan2A) is the required .....

Here, LHS =(1 - cos π/8)(1 - cos 3π/8)(1 - cos 5π/8)(1 - cos 7π/8) = (1 - cos π/8)(1 - cos 3π/8)(1 + cos 3π/8)(1 + cos π/8), [since cos(π - A) = -cos A.] = (1 - cos² π/8)(1 - cos² 3π/8) = (1 - .....

Here, LHS = sin4b + cos4b  = (sin2b)2 + (cos2b)2 = (sin2b + cos2b) - 2sin2b cos2b    [ Since, a2 .....

We Know Sin (A + B) = sin A cos B + cos A sin B Putting B = 4A we get sin 5A = sin ( A + 4 A)            = sin A cos 4A + cos A sin 4A            = sin A (1- 2 Sin2 2A) + cos A .2Sin 2A .....

Here, L.H.S. is in the form of (a - b)2, So, you can write the expression as = (sin2A/2) - 2 (sin A/2) (cos A/2) + (cos2A/2) = (sin2A/2) + (cos2A/2) - 2 (sin A/2) (cos A/2)  = 1 - sinA   [Since, sin A = 2 (sin A/2) .....

Here, tanA= m and tanB= 1m We know that, tan (A+B) = tan⁡tanA+tan⁡tanB1- tan⁡tanA tan⁡tanB                                           = .....

Here, LHS = sin2A cos2B - sin2B cos2A  = (1 - cos2A)(2cos2B - 1) - (1 - cos2B)(2cos2A - 1)   [ Since, sin2X = 1 - cos2X and cos2X = 2cos2X - 1).] Opening bracket we get = 2cos2B - 1 - 2cos2A.cos2B + cos2A -{2cos2A - 1 - .....

Here,     Sin A = 45 We have, Sin A = 3sin A3 -4 sin3A3           =3×45 - 4 ×(45)3           = 125 – 4×64125          = 125 – .....

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

Here,  LHS = 8 ( 1 + sin πc 8) ( 1 + sin 3πc 8) ( 1 - sin 5πc 8) ( 1 - sin 7πc 8)          =  8 1 + sin πc 8( 1 + sin⁡3πc 8 ) [1 - sin⁡ ( πc- 3πc/8)][ 1 - sin⁡( πc .....