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.