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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Question 1 :

To Prove :

$\sf\dfrac{sec\theta+cos\theta}{sec\theta-cos\theta}=\dfrac{1+cos^2{\theta}}{sin^2{\theta}}$

LHS :

$\sf=\dfrac{sec\theta+cos\theta}{sec\theta-cos\theta}$

$\sf=\dfrac{\frac{1}{cos\theta}+cos\theta}{\frac{1}{cos\theta}-cos\theta}$

$\sf=\dfrac{\frac{1+cos^2{\theta}}{cos\theta}}{\frac{1-cos^2{\theta}}{cos\theta}}$

$\sf=\dfrac{\frac{1+cos^2{\theta}}{\cancel{cos\theta}}}{\frac{1-cos^2{\theta}}{\cancel{cos\theta}}}$

$\sf=\dfrac{1+cos^2{\theta}}{1-cos^2{\theta}}$

• Using 1 - cos²A = sin²A,

$\sf=\dfrac{1+cos^2{\theta}}{sin^2{\theta}}$

RHS :

$\sf=\dfrac{1+cos^2{\theta}}{sin^2{\theta}}$

$\bf\therefore$ LHS = RHS.

• Hence proved.

Question 2 :

To Prove :

$\sf\dfrac{sin^3{\theta}-cos^3{\theta}}{sin\theta-cos\theta}=1+sin\theta cos\theta$

LHS :

$\sf=\dfrac{sin^3{\theta}-cos^3{\theta}}{sin\theta-cos\theta}$

• Using a³ - b³ = (a - b)(a² + ab + b²)

$\sf=\dfrac{(sin\theta-cos\theta)(sin^2{\theta}+sin\theta cos\theta+cos^2{\theta}}{sin\theta-cos\theta}$

$\sf=\dfrac{\cancel{(sin\theta-cos\theta)}(sin^2{\theta}+sin\theta cos\theta+cos^2{\theta})}{\cancel{(sin\theta-cos\theta)}}$

$\sf=sin^2{\theta}+sin\theta cos\theta+cos^2{\theta}$

$\sf=sin^2{\theta}+cos^2{\theta}+sin\theta cos\theta$

• Using cos²A + sin²A = 1,

$\sf=1+sin\theta cos\theta$

RHS :

$\sf=1+sin\theta cos\theta$

$\bf\therefore$ LHS = RHS.

• Hence proved.