Soln:

Here, (sinA + sinB) (sinA - sinB)

= sinA (sinA - sinB) + sinB (sinA - sinB)

= sin2A - sinA.sinB + sinA.sinB - sin2B

= sin2A - sin2B

Soln:

Here, ( 1 - cosθ) ( 1 + cosθ)

= 1 - cos2θ

= sin2θ. Ans

Soln:

Here, ( 1+ tanθ) ( 1 - tanθ) ( 1 + tan2θ)

= (1 - tan2θ) (1 + tan2θ) = 1 - tan4θ Ans.

Soln:

cos2A - sin2A = ( cosA - sinA) (cosA + sinA). Ans.

Soln:

L. H. S. = (1 + tan2A) cos2A

= sec2A. cos2A

= \(\frac{1}{cos^2A}\). cos2A

= 1 = RHS Proved.

Soln:

LHS =\(\frac{1}{cos^2A}\) - \(\frac{1}{cot^2A}\) = sec2A -- tan2A = 1 + tan2A - tan2A

=1 = RHS proved

Soln:

L.H.S. =\(\frac{secA}{cosA}\) - \(\frac{tanA}{cotA}\) = secA. \(\frac{1}{cosA}\) - tanA. \(\frac{1}{cotA}\)

= secA. secA - tanA. tanA

= sec2A - tan2A = 1 = R. H. S. Proved

Soln:

Here,\(\frac{1-tanA}{1+tanA}\) =\(\frac{\frac{1}{tanA}-\frac{tanA}{taanA}}{\frac{1}{tanA}+\frac{tanA}{tanA}}\)

= \(\frac{cotA - 1}{cotA + 1}\) RHS proved

Soln:

L.H.S = ( tanθ + secθ)2

= (\(\frac{sinθ}{cosθ}\) + \(\frac{1}{cosθ}\))2

= (\(\frac{1 + sinθ}{cosθ}\))2

= \(\frac{(1 + sinθ)( 1 + sinθ)}{cos^2θ}\)

=\(\frac{(1 + sinθ)(1 + sinθ)}{1 - sin^2θ}\)

= \(\frac{1 + sinθ}{1 - sinθ}\) = RHS Proved.

Soln:

LHS. =\(\frac{1}{1 - cosA}\) - \(\frac{1}{1 + cosA}\) = \(\frac{1 + cosA -(1 - cosA)}{(1 - cosA) ( 1 + cosA)}\)

= \(\frac{1 + cosa - 1 + cosA}{1 - cos^2 A}\)

= \(\frac{2cosA}{sin^2 A}\) =\(\frac{2cosA}{sinA}\) .\(\frac{1}{sinA}\)

= 2cotA cosecA = RHS Proved.

Soln:

Taking LHS,
=\(\frac{cosx}{1 - sinx}\) + \(\frac{cosx}{1 + sinx)}\)

= \(\frac{cosx(1 + sinx) + cosx(1 - sinx)}{(1 - sinx) (1 + sinx)}\)

= \(\frac{cosx + cosx.sinx + cosx - cosx.sinx}{1 - sin^2 x}\)

= \(\frac{2cosx}{cos^2x}\) =\(\frac{2}{cosx}\) = 2secx = RHS proved. Ans

Soln:
Taking LHS,
=sin4θ + cos4θ = (sin2θ)2 + (cos2θ)2

=(sin2θ + cos2θ)2 - 2sin2θ cos2θ

= 1 - 2sin2θ cos2θ = RHS Proved

Soln:

LHS = ( 1 + sinA + cosA)2

={(1 + sinA) +cosA}2

= (1 + sinA)2 + 2(1 + sinA) cosA + cos2A

=(1 + sinA)2 + 2( 1 + sinA) cosA + cos2A

= ( 1+ sinA)2+ 2(1 + sinA) cosA + 1 - sin2A

= ( 1+ sinA)2 + 2(1 + sinA) cosA + (1 + sinA) (1 - sinA)

=(1 + sinA) ( 1 + sinA + 2cosA + 1 -sinA)

= ( 1 + sinA) ( 2 + 2 cosA)

= 2(1 + sinA) (1 + cosA) = RHS Proved

Soln:

LHS =\(\frac{secA - tanA}{secA + tanA}\) =\(\frac{(secA - tanA)}{secA + tanA}\)× \(\frac{secA - tanA}{secA - tanA}\)

= \(\frac{(secA - tanA)^2}{sec^2 A -tan^2A}\)

=\(\frac{sec^2 A - 2secA tanA + tan^2A}{1}\)

= 1 + tan2A - 2secA tanA + tan2A

= 1 - 2secA tanA + 2 tan2A = RHS proved.

Soln:

RHS = 1 + 2cotA + cot2A + 1 + 2 tanA + tan2A

= 1 + cot2A + 1 + tan2A + 2 (cotA + tanA)2

= 1 + cot2A + 1 + tan2A + 2 (cotA + tanA)

= cosec2A + sec2A + 2(\(\frac{cosA}{sinA}\) + \(\frac{sinA}{cosA}\))

=coscec2A +sec2A + 2(\(\frac{cos^2A + sin^2A}{sinA cosA}\))

= cosec2A + sec2A + 2. \(\frac{1}{sinA cosA}\)

= cosec2A + sec2A + 2coosecA secA

= (secaA + cosecA)2 =LHS proved

soln: Here, sec4θ - cosec4θ

= (sec2θ)2 - (cosec2θ)2

= ( sec2θ - cosec2θ) (sec2θ + cosec2θ)

= ( secθ - cosecθ ) (secθ - cosecθ ) (sec2θ + cosec2θ)

soln: here, sin2x + 3sinx + 2

= sin2x + 2sinx + sinx + 2

= sinx (sinx + 2) +1 (sinx + 2)

= (sinx + 2 ) (sinx + 1)

soln: L.H.S =(1 - cos2 A ) (1 + cot2 A ) = 1 ∴ 1 + cot2θ

= sin2 A . cosec2 A = sin2 A× \(\frac{1}{sin^2}A\) = cosec2 θ

= 1 = R.H.S proved. and 1 - cos2θ = sin2θ

soln: L.H.S=tanθ . \(\sqrt{1} {sin^2θ}\) = tanθ \(\sqrt{cos^2θ}\)

= tanθ . cosθ = \(\frac{sinθ}{cosθ}\).cosθ=sinθ= R.H.S. proved

soln: L.H.S = tan2 A - sin2 A = \(\frac{sin^2A}{cos^2A}\) - sin2 A

= \(\frac{sin^2 A- sin^2 A cos^2 A}{cos^2}\)=\(\frac{sin^2 (1 - cos^2 A)}{cos^2}\)

= sin2A . \(\frac{sin^2A}{cos^2A}\)= sin2 A. tan2 A

= R.H.S. proved.

soln: Given,θ = 300

L.H.S. = sin3θ = sin 3× 300 = sin900 = 1

R.H.S. = 3sinθ - 4sin3θ = 3sin300- 4sin3300

= 3×\(\frac{1}{2}\)-4 \(\begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}\)3 = \(\frac{3}{2}\)- \(\frac{4}{8}\)

= \(\frac{3}{2}\)- \(\frac{1}{2}\)= \(\frac{3-1}{2}\)=\(\frac{2}{2}\)=1

∴ L.H.S.= R.H.S. proved.

Soln:

LHS =\(\frac{cosA - sinA + 1}{cosA + sinA - 1}\)

=\(\frac{\frac{cosA - sinA + 1}{sinA}}{\frac{cosA + sinA - 1}{sinA}}\)

=\(\frac{cotA - 1 + cosecA}{cotA + 1 - cosecA}\)

= \(\frac{(cotA + cosecA) - (cosec^2 A - cot^2 A)}{cotA - cosecA + 1}\)

=\(\frac{1 (cotA + cosecA) - (cosecA - cotA) (cosecA + cotA) }{cotA - cosecA + 1}\)

= \(\frac{(cosecA + cotA) (1 - cosecA + cotA)}{cotA - cosecA + 1}\)

= cosecA + cotA =\(\frac{1}{sinA}\) + \(\frac{cosA}{sinA}\)

= \(\frac{1 + cosA}{sinA }\) =RHS proved.

soln; L.H.S=\(\frac{1}{secA-tanA}\)-\(\frac{1}{cosA}\)

= \(\frac{sec^2A-tan^2A}{secA-tanA}\)-\(\frac{1}{cosA}\)

= \(\frac{(secA+tanA)(secA-tanA}{(secA-tanA}\)-secA

= secA - secA + tanA = tanA

∴ L.H.S.= R.H.S. proved.

R.H.S = \(\frac{1}{cosA}\)-\(\frac{1}{secA + tanA}\)

= secA-\(\frac{sec^2+tan^2}{secA + tanA}\)

= sec A-\(\frac{(secA+tanA)(secA-tanA)}{(secA+tanA)}\)

= secA - secA + tanA=tanA

∴ L.H.S=R.H.S. proved.

soln:L.H.S. = cosec4A (1-cos4A)

= cosec4A(1-cos2 A) (1+cos2 A)= cosec4θ[1-(cos2A)2]

= cosec2 A. cosec2 A. sin2 A (1+cos2A)

= cosec2A.\(\frac{1}{sin^2A}\).sin2A(1+cos2A)

= cosec2A (1+cos2A)

= cosec2A+cosec2A cos2A= cosec2A + \(\frac{cos^2A}{sin^2A}\)

= cosec2A + cot2 A = 1 + cot2A+cot2A

= 1+2 cot2A= R.H.S. proved.

soln: given,∝=00,β=300

L.H.S.= sin(∝ + β) = sin (00 + 300)= sin300=\(\frac{1}{2}\)

R.H.S = sin∝ cosβ + cos∝ sinβ

= sin00 cos00 + cos00 sin00

= 0×\(\frac{\sqrt{3}}{2}\)+ 1 \(\begin{bmatrix} 1 \\ 2 \end{bmatrix}\) = 0 +\(\frac{1}{2}\)=\(\frac{1}{2}\)

∴ L.H.S = R.H.S. proved.