**sin**^{2 }^{π}/_{6}+ cos^{2}^{π}/_{3}– tan^{2}^{π}/_{4}= –^{1}/_{2}

Solution:

L.H.S = sin^{2} ^{π}/_{6} + cos^{2 π}/_{3} – tan^{2} ^{π}/_{4}

= (^{1}/_{2})^{2} + (^{1}/_{2})^{2} – (1)^{2}

= ^{1}/_{4 }+ ^{1}/_{4 }– 1 = –^{1}/_{2}

= R.H.S

**Prove that 2sin**^{2}^{π}/_{6}+ coses^{2 7π}/_{6}cos^{2 π}/_{3}=^{3}/_{2}

Solution:

L.H.S = 2sin^{2} ^{π}/_{6} + coses^{2 7π}/_{6} cos^{2 π}/_{3}

= 2(^{1}/_{2})^{2} + cosec^{2 }(π + ^{π}/_{6})(^{1}/_{2})^{2}

= 2(^{1}/_{4}) + (-cosec ^{π}/_{6})^{2}(^{1}/_{4})

= ^{1}/_{2} + (-2)^{2}(^{1}/_{4})

= ^{1}/_{2} + (^{4}/_{4}) = = ^{1}/_{2} + 1 = ^{3}/_{2}

= R.H.S

**Prove that cot**^{2 π}/_{6}+ cosec^{5π}/_{6}+ 3tan^{2}^{π}/_{6}

Solution:

L.H.S = cot^{2 π}/_{6} + cosec ^{5π}/_{6} + 3tan^{2} ^{π}/_{6}

= (√3)^{2} + cosec(π-^{ π}/_{6}) +3(^{1}/_{√3})^{2}

= 3 + cosec ^{π}/_{6} + 3 x ^{1}/_{3}

= 3 + 2 + 1

= 6

= R.H.S

**Prove that 2 sin**^{2}(^{3}^{π}/_{4}) + 2cos^{2}(^{ π}/_{4}) +2sec^{2}(^{π}/_{3}) = 10

Solution:

L.H.S = 2 sin^{2} (^{3}^{π}/_{4}) + 2cos^{2}(^{ π}/_{4}) +2sec^{2}(^{π}/_{3})

= 2{sin(π – ^{π}/_{4})}^{2} + 2(^{1}/_{√2})^{2} + 2(2)^{2}

=2{sin^{π}/_{4}}^{2} + 2 x ^{1}/_{2} + 8

= 2(^{1}/_{√2})^{2} + 1+ 8

= 1 + 1 + 8

= 10

= R.H.S

**Find the value of**

**(i) sin 75˚**

**(ii)tan15˚**

Solution:

(i) sin 75˚

= sin 75˚ cos 30˚ + cos 45˚ sin 30˚

[sin(x + y) = sin x cos y + cos x sin y]

= (^{1}/_{√2}) x(^{√3}/_{2}) + (^{1}/_{√2}) x(^{1}/_{2})

= (^{√3}/_{2√2})+(^{1}/_{2√2})

=^{√3+1}/_{2√2}

(ii) tan15˚

= tan(45˚ – 30˚)

= ^{tan45}˚^{-tan30}˚/_{1+tan45˚-tan30˚}

= ^{[1-(1/√3)]}/_{[1+1(1/}_{ √3)] }

= ^{[(√3-1)/√3]}/_{[(√3+1)/√3]}

= ^{(√3-1)}/_{(√3+1)}

= (√3-1)^{2}/[(√3+1)(√3-1)]

= ^{3+1-2√3}/_{(√3)}^{2}_{-(1)}^{2}

=^{ (4 – 2√3)}/_{3-1} = 2 – √3

**Prove that cos (**^{π}/_{4}– x) cos(^{π}/_{4}– y) – sin(^{π}/_{4}– x)sin(^{π}/_{4}– y) = sin(x+y)

Solution:

L.H.S = cos (^{π}/_{4} – x) cos(^{π}/_{4} – y) – sin(^{π}/_{4} – x)sin(^{π}/_{4} – y)

= ^{1}/_{2} [2cos (^{π}/_{4} – x) cos(^{π}/_{4} – y)] +^{1}/_{2}[ –2 sin(^{π}/_{4} – x)sin(^{π}/_{4} – y)]

=^{1}/_{2}[cos{(^{ π}/_{4} – x)+(^{ π}/_{4} – y)}+cos{(^{ π}/_{4} – x)-(^{ π}/_{4} – y)}] + ^{1}/_{2}[cos{(^{ π}/_{4} – x)+(^{ π}/_{4} – y)} – cos{(^{ π}/_{4} – x)-(^{ π}/_{4} – y)}]

[2cosAcosB = cos(A+B)+cos(A-B)

-2sinAsinB = cos(A+B)-cos(A-B)]

= 2 x ^{1}/_{2}[cos{(^{ π}/_{4} – x)+(^{ π}/_{4} – y)}]

=cos [^{π}/_{2} –(x + y)]

= sin(x+y)

= R.H.S

**Prove that**^{tan[(π/4) +x]}/_{tan[(π/4) -x]}= [^{1+tanx}/_{1-tanx}]^{2}

Solution:

tan(A+B) = ^{tanA+tanB}/_{1-tanAtanB} and tan(A-B) = ^{tanA-tanB}/_{1+tanAtanB}

L.H.S = ^{tan[(π/4) +x]}/ _{tan[(π/4) -x]} = ^{[tan(π/4) +tanx]/[1-tan (π/4)tanx]}/ _{[tan(π/4) -tanx]} _{[1+tan (π/4)tanx]}

= ^{[(1+tanx)/(1-tanx)]}/ _{[(1-tanx)/(1+tanx)]}

= [^{(1+tanx)}/_{(1-tanx)}]^{2}

= R.H.S

**Prove that**^{cos(π+x)cos(-x)}/_{sin(π-x)cos((π/2)+x)}= cot^{2}x

Solution:

L.H.S = ^{cos(π+x)xos(-x)}/_{sin(π-x)cos((π/2)+x)}

= ^{[-cosx][cosx]}/_{[sinx][-sinx]}

= [-cos^{2}x]/[-sin^{2}x]

=cot^{2}x

= R.H.S

**cos(**^{3π}/_{2 }+ x)cos(2π+x)[cot (^{3π}/_{2 }– x)cot(2π+x)] = 1

Solution:

L.H.S = cos(^{3π}/_{2 }+ x)cos(2π+x)[cot (^{3π}/_{2 }– x)cot(2π+x)]

= sinx cosx[tanx + cotx]

= sinx cosx (^{sinx}/_{cosx} + ^{cosx}/_{sinx})

=(sinx cosx)[(sin^{2}x + cos^{2}x)/(sinx cosx)]

=1

=R.H.S

**Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x**

Solution:

L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x

= ^{1}/_{2}(2sin(n+1)x sin(n+2)x + 2cos(n+1)x cos(n+2)x]

=^{1}/_{2}[cos{(n+1)x-(n+2)x}-cos{(n+1)x+(n+2)x}+cos{(n+1)x+(n+2)x}+cos{(n+1)x-(n-2)x}]

[-2sinAsinB = cos(A+B)-cos(A-B)

2cosAcosB = cos(A+B)+cos(A-B)]

=^{1}/_{2} x 2cos{(n+1)x-(n+2)x}

=cos(-x)

=cos x

=R.H.S

**Prove that cos(**^{3}^{π}/_{4}+ x)-cos(^{3}^{π}/_{4}+ x) = -√2.sinx

Solution:

We know that cosA – cosB = -2sin(^{A+B}/_{2}).sin(^{A-B}/_{2})

L.H.S = cos(^{3}^{π}/_{4} + x)-cos(^{3}^{π}/_{4} + x)

= -2sin[^{(3}^{π/4 + x)+(3}^{π/4 – x)}/_{2}].sin[^{(3}^{π/4 + x)-(3}^{π/4 – x)}/_{2}]

= -2 sin(^{3}^{π}/_{4}) sinx

= -2 sin(π – ^{π}/_{4}) sinx

= -2 sin^{ π}/_{4}.sinx

= -√2.sinx

**Prove that sin**^{2}6x – sin^{2}4x = sin 2x sin 10x

Solution:

We know that,

sinA+sinB = 2sin[^{A+B}/_{2}]cos[^{A-B}/_{2}] ;

sinA-sinB = 2cos[^{A+B}/_{2}]sin[^{A-B}/_{2}]

L.H.S. = sin^{2}6x – sin^{2}4x

= (sin6x+sin4x)(sin6x-sin4x)

=[ 2sin[^{6x+4x}/_{2}]cos[^{6x-4x}/_{2}] [2cos[^{6x+4x}/_{2}]sin[^{6x-4x}/_{2}]

= (2sin5xcosx)(2cos5xsinx)

= (2sin5xcos5x)(2sinxcosx)

= sin10x sin2x

= R.H.S

**Prove that cos**^{2}2x – cos^{2}6x = sin 4x sin 8x

Solution:

We know that

cosA+cosB = 2cos[^{A+B}/_{2}]cos[^{A-B}/_{2}];

cosA – cosB = -2sin[^{A+B}/_{2}]sin[^{A-B}/_{2}];

L.H.S = cos^{2}2x – cos^{2}6x

= (cos 2x + cos 6x)(cos 2x – cos 6x)

= {2cos[^{2x+6x}/_{2}]cos[^{2x-6x}/_{2}]}{ -2sin[^{2x+6x}/_{2}]sin[^{2x-6x}/_{2}]

= [2cos4xcos(-2x)][-2sin4xsin(-2x)]

=[2cos4x cos2x][-2 sin4x (-sin2x)]

=[2 cos 4x cos2x][-2 sin4x(-sin2x)]

= (2 sin 4x cos 4x) (2 sin 2x cos 2x)

= sin 8x sin 4x

= R.H.S.

**14: Prove that sin 2x + 2sin 4x + sin 6x = 4cos ^{2}x sin 4x**

Solution:

L.H.S. = sin 2x + 2 sin 4x + sin 6x

= [sin 2x + sin 6x] + 2 sin 4x

=[2sin(^{2x+6x}/_{2})(^{2x-6x}/_{2})]+2sin4x

[sinA+sinB = 2sin(^{A+B}/_{2})cos(^{A-B}/_{2})]

= 2 sin4x cos(– 2x) + 2 sin4x

= 2 sin4x cos2x + 2 sin4x

= 2 sin4x (cos2x + 1)

= 2 sin4x (2 cos^{2}x – 1 + 1)

= 2 sin4x (2 cos^{2}x)

= 4cos^{2}x sin 4x

= R.H.S.

**15: Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)**

Solution:

L.H.S = cot 4x (sin 5x + sin 3x)

=^{cot4x}/_{sin4x} [2sin(^{5x+3x}/_{2})cos(^{5x-3x}/_{2})]

[sinA + sinB = 2sin(^{A+B}/_{2})cos(^{A-B}/_{2})]

=[^{cos4x}/_{sin4x}_{ ][2sin4xcosx]}

= 2 cos 4x cos x

R.H.S. = cot x (sin 5x – sin 3x)

=^{cosx}/_{sinx} [2cos(^{5x+3x}/_{2})sin(^{5x-3x}/_{2})

[sinA – sinB = 2cos(^{A+B}/_{2})sin(^{A-B}/_{2})]

= ^{cosx}/_{sinx} [2cos4x sinx]

= 2 cos 4x. cos x

L.H.S. = R.H.S.

**Prove that**^{cos9x – cos5x}/_{sin17x – sin3x}= –^{sin2x}/_{cos10x}

Solution:

We know that,

cosA – cosB = -2sin[^{A+B}/_{2}]sin[^{A-B}/_{2}];

sinA – sinB = 2cos[^{A+B}/_{2}]sin[^{A-B}/_{2}];

L.H.S = ^{cos9x – cos5x}/_{sin17x – sin3x}

={-2sin[^{9x+5x}/_{2}]sin[^{9x-5x}/_{2}]} / {2cos[^{17x+3x}/_{2}]sin[^{17x-3x}/_{2}]}

= ^{{-2sin7x.sin2x}}/_{{2cos10x.sin7x}}

= – ^{sin2x}/_{cos10x}

= R.H.S.

**Prove that**^{sin5x+sin3x}/_{cos5x+cos3x}= tan4x

Solution:

We know that,

sinA+sinB = 2sin(^{A+B}/_{2})cos(^{A-B}/_{2});

cosA+cosB = 2cos(^{A+B}/_{2})cos(^{A-B}/_{2});

L.H.S. = ^{sin5x+sin3x}/_{cos5x+cos3x}

= {2sin(^{5x+3x}/_{2})cos(^{5x-3x}/_{2})} / {2cos(^{5x+3x}/_{2})cos(^{5x-3x}/_{2})

= ^{2sin4x.cosx}/_{2cos4x.cosx}

= ^{sin4x}/_{cos4x}

= tanx

= R.H.S.

**Prove that**^{sinx-siny}/_{cosx+cosy}= tan^{x-y}/_{2}

Solution:

We know that

sinA – sinB = 2cos(^{A+B}/_{2})sin(^{A-B}/_{2});

cosA+cosB = 2cos(^{A+B}/_{2})cos(^{A-B}/_{2});

L.H.S. = ^{sinx-siny}/_{cosx+cosy}

= {2cos(^{x-y}/_{2})sin(^{x-y}/_{2})} / {2cos(^{x+y}/_{2})cos(^{x-y}/_{2})}

= {sin(^{x-y}/_{2})} / {cos(^{x-y}/_{2})}

= tan (^{x-y}/_{2})

= R.H.S.

**Prove that**^{sinx+sin3x}/_{cosx+cos3x}= tan 2x

Solution:

We know that

sinA+sinB = 2sin(^{A+B}/_{2})cos(^{A-B}/_{2});

cosA+cosB = 2cos(^{A+B}/_{2})cos(^{A-B}/_{2});

L.H.S. = ^{sinx+sin3x}/_{cosx+cos3x}

= {2sin(^{x+3x}/_{2})cos(^{x-3x}/_{2})} / {2cos(^{x+3x}/_{2})cos(^{x-3x}/_{2})}

= ^{sin2x}/_{cos2x}

= tan2x

=R.H.S

**Prove that**^{{sinx-sin3x}}/{sin^{2}x-cos^{2}x} = 2sinx

Solution:

We know that,

sinA – sinB = 2cos(^{A+B}/_{2})sin(^{A-B}/_{2});

cos^{2}A – sin^{2}A = cos 2A

L.H.S = ^{{sinx-sin3x}}/{sin^{2}x-cos^{2}x}

= {2cos(^{x+3x}/_{2})sin(^{x-3x}/_{2})} / {-cos2x}

=^{2cos2xsin(-x)}/_{-cos2x}

=-2(-sinx)

=2sinx

=R.H.S.

**Prove that**^{cos4x+cos3x+cos2x}/_{sin4x+sin3x+sin2x}= cot3x

Solution:

L.H.S. = ^{cos4x+cos3x+cos2x}/_{sin4x+sin3x+sin2x}

= ^{(cos4x+cos2x)+cos3x}/_{(sin4x+sin2x)+sin3x}

= {2cos(^{4x+2x}/_{2})cos(^{4x-2x}/_{2})+cos3x} / {2sin(^{4x+2x}/_{2})cos(^{4x-2x}/_{2})+sin3x}

**[cosA+cosB = 2cos( ^{A+B}/_{2})cos(^{A-B}/_{2});**

**sinA+sinB = 2sin( ^{A+B}/_{2})cos(^{A-B}/_{2})]**

=^{{2cos3x cosx + cos3x}}/_{{2sin3x cosx + sin3x}}

=^{cos3x(2cosx+1)}/_{sin3x(2cosx+1)}

=cot3x

=R.H.S.

**Prove that cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1**

Solution:

L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – cot (2x + x) (cot 2x + cot x)

= cotx cot2x – [^{cot2xcotx-1}/_{cotx+cot2x}](cot2x+cotx)

[cot(A+B) = ^{cotAcotB-1}/_{cotA+cotB}]

= cot x cot 2x – (cot2x cotx – 1)

=1

=R.H.S.

**Prove that tan4x = {4tanx(1-tan**^{2}x)} / {1-6tan^{2}x +tan^{4}x}

Solution:

We know that,

tan2A = ^{2tanA}/_{1-tan}^{2}_{A}

L.H.S. =tan4x = tan 2(2x)

^{ }

**Prove that: cos 4x = 1 – 8sin**^{2}x cos^{2}x

Solution:

L.H.S. = cos 4x

= cos 2(2x)

= 1 – 2 sin^{2} 2x

[cos 2A = 1 – 2 sin2 A]

= 1 – 2(2 sin x cos x)^{2}

[sin2A = 2sin A cosA]

= 1 – 8 sin^{2}x cos^{2}x

= R.H.S.

**25: Prove that: cos 6x = 32 cos ^{6}x – 48 cos^{4}x + 18 cos^{2}x – 1**

Solution:

L.H.S. = cos 6x

= cos 3(2x)

= 4 cos^{3} 2x – 3 cos 2x

[cos 3A = 4 cos^{3}A – 3 cos A]

= 4 [(2 cos^{2}x – 1)^{3} – 3 (2 cos^{2}x – 1)

[cos 2x = 2 cos^{2}x – 1]

= 4 [(2 cos^{2}x)^{3} – (1)^{3} – 3 (2 cos2 x) 2 + 3 (2 cos^{2} x)] – 6cos^{2} x + 3

= 4 [8cos^{6}x – 1 – 12 cos^{4}x + 6 cos^{2}x] – 6 cos^{2}x + 3

= 32 cos^{6}x – 4 – 48 cos^{4}x + 24 cos^{2} x – 6 cos^{2}x + 3

= 32 cos^{6}x – 48 cos^{4}x + 18 cos^{2}x – 1

= R.H.S.

^{ }