Solution

(sinθ + cosθ) (sinθ + cosθ)

= sin$$\theta$$(sinθ + cosθ) (sinθ + cosθ) + cos$$\theta$$(sinθ + cosθ) (sinθ + cosθ)

= sin2$$\theta$$ + sin$$\theta$$ × cos$$\theta$$ + cos$$\theta$$ × sin$$\theta$$ + cos2$$\theta$$

= sin2$$\theta$$ + sin$$\theta$$ × cos$$\theta$$ + cos$$\theta$$ × sin$$\theta$$ + cos2$$\theta$$

= sin2$$\theta$$ + 2 sin$$\theta$$ × cos$$\theta$$ + cos2$$\theta$$

Solution

(1- tan$$\theta$$) (1+tan$$\theta$$) (1+tan2$$\theta$$) (1+tan4$$\theta$$)

= (1 - tan2$$\theta$$)(1 + tan2$$\theta$$)(1 + tan4$$\theta$$)

= (1 - tan4$$\theta$$)(1 + tan4$$\theta$$)

1 - tan8$$\theta$$

Solution

$$\frac{1}{1 + cosθ}$$ + $$\frac{1}{1-cosθ}$$

=  $$\frac{1}{1 + cosθ}$$ + $$\frac{1}{1-cosθ}$$

= $$\frac{1-cosθ +1+cosθ}{(1-cosθ)(1-cosθ)}$$

= $$\frac{2}{1-cosθ}$$ = $$\frac{2}{sin^2θ}$$ =  2 cosec2θ

Solution

= $$\frac{1}{secA - tanA}$$ - $$\frac{1}{secA + tanA}$$

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

= $$\frac{secA + tanA - secA + tanA}{1}$$

= 2 tanA

L.H.S = $$\frac{1 - sin\alpha}{cos\alpha}$$

= $$\frac{1 - sin\alpha}{cos\alpha}$$ × $$\frac{1 + sin\alpha}{1 + sin\alpha}$$

= $$\frac{1 - sin\alpha}{cos\alpha(1 + sin\alpha)}$$

= $$\frac{cos\alpha}{(1 + sin\alpha)cos\alpha}$$

= $$\frac{cos\alpha}{1 +sin\alpha}$$ = R.H.S

$$\therefore$$ LHS = RHS proved.

Solution'

L.H.S = $$\frac{secA - tanA + 1}{secA - tanA - 1}$$

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

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

= $$\frac{(secA - tanA)(1 + secA + tanA)}{(secA - tanA)(1 - secA - tanA)}$$

= $$\frac{1 + secA + tanA}{1 - secA - tanA}$$ = R.H.S

$$\therefore$$ L.H.S = R.H.S proved.

Solution

Here, $$\frac{1 + 3sin\theta - 4sin^3\theta}{1 - sin\theta}$$

= $$\frac{1 - sin\theta + 4sin\theta - 4sin^2\theta + 4sin^2\theta - 4sin^3\theta}{1 - sin\theta}$$

= $$\frac{1 (1 - sin\theta) + 4 sin\theta(1 - sin\theta) + 4 sin^2\theta(1 - sin\theta)}{1 - sin\theta}$$

= $$\frac{(1 - sin\theta)(1 +4sin\theta + 4sin^2\theta)}{1 - sin\theta}$$

= (1)2 + 2×1×2sin\)\theta\) + (2sin$$\theta$$)2

= (1 + 2sin$$\theta$$)2 = RHS

$$\therefore$$ LHS = RHS proved.