Question

Prove the following trigonometric identities:

Prove that:

(i) $\sqrt{\frac{sec\theta -1}{sec\theta +1}}+\sqrt{\frac{sec\theta +1}{sec\theta -1}}=2cosec\theta$

(ii) $\sqrt{\frac{1+sin\theta }{1-sin\theta }}+\sqrt{\frac{1-sin\theta }{1+sin\theta }}=2cosec\theta$

(iii) $\sqrt{\frac{1+cos\theta }{1-cos\theta }}+\sqrt{\frac{1-cos\theta }{1+cos\theta }}=2cosec\theta$

(iv) $\frac{sec\theta -1}{sec\theta +1}={\left(\frac{sin\theta }{1+cos\theta }\right)}^2$

Answer 1

(i) √(secΘ−1)(secΘ+1)+√(secΘ+1)(secΘ−1)(secΘ−1)(secΘ+1)−−−−−−√+(secΘ+1)(secΘ−1)−−−−−−√ = 2cosec ΘΘ

Ans:

To prove,

= √(secΘ−1)(secΘ+1)+√(secΘ+1)(secΘ−1)(secΘ−1)(secΘ+1)−−−−−−√+(secΘ+1)(secΘ−1)−−−−−−√ = 2cosec ΘΘ

Considering left hand side (LHS),

Rationalize the numerator and denominator.

= √(secΘ−1)(secΘ−1)(secΘ+1)(secΘ−1)+√(secΘ+1)(secΘ+1)(secΘ−1)(secΘ+1)(secΘ−1)(secΘ−1)(secΘ+1)(secΘ−1)−−−−−−−−−−−−√+(secΘ+1)(secΘ+1)(secΘ−1)(secΘ+1)−−−−−−−−−−−−√

= √(secΘ−1)2(sec2Θ−1)+√(secΘ+1)2(sec2Θ−1)(secΘ−1)2(sec2Θ−1)−−−−−−−√+(secΘ+1)2(sec2Θ−1)−−−−−−−√

= √(secΘ−1)2tan2Θ+√(secΘ+1)2tan2Θ(secΘ−1)2tan2Θ−−−−−−−√+(secΘ+1)2tan2Θ−−−−−−−√

= (secΘ−1)tanΘ+(secΘ+1)tanΘ(secΘ−1)tanΘ+(secΘ+1)tanΘ

= (secΘ−1+secΘ+1)tanΘ(secΘ−1+secΘ+1)tanΘ

= (2cosΘ)cosΘsinΘ(2cosΘ)cosΘsinΘ

= 2sinΘ2sinΘ

= 2cosec ΘΘ

Therefore, LHS = RHS

Hence proved

(ii) √(1+sinΘ)(1−sinΘ)+√(1−sinΘ)(1+sinΘ)(1+sinΘ)(1−sinΘ)−−−−−−√+(1−sinΘ)(1+sinΘ)−−−−−−√ = 2sec ΘΘ

Ans:

To prove,

= √(1+sinΘ)(1−sinΘ)+√(1−sinΘ)(1+sinΘ)(1+sinΘ)(1−sinΘ)−−−−−−√+(1−sinΘ)(1+sinΘ)−−−−−−√ = 2sec ΘΘ

Considering left hand side (LHS),

Rationalize the numerator and denominator.

= √(1+sinΘ)(1+sinΘ)(1−sinΘ)(1+sinΘ)+√(1−sinΘ)(1−sinΘ)(1+sinΘ)(1−sinΘ)(1+sinΘ)(1+sinΘ)(1−sinΘ)(1+sinΘ)−−−−−−−−−−−−√+(1−sinΘ)(1−sinΘ)(1+sinΘ)(1−sinΘ)−−−−−−−−−−−−√

= √(1+sinΘ)2(1−sin2Θ)+√(1−sinΘ)2(1−sin2Θ)(1+sinΘ)2(1−sin2Θ)−−−−−−−√+(1−sinΘ)2(1−sin2Θ)−−−−−−−√

= √(1+sinΘ)2(cos2Θ)+√(1−sinΘ)2(cos2Θ)(1+sinΘ)2(cos2Θ)−−−−−−−√+(1−sinΘ)2(cos2Θ)−−−−−−−√

= (1+sinΘ)(cosΘ)+(1−sinΘ)(cosΘ)(1+sinΘ)(cosΘ)+(1−sinΘ)(cosΘ)

= (1+sinΘ+1−sinΘ)(cosΘ)(1+sinΘ+1−sinΘ)(cosΘ)

= (2)(cosΘ)(2)(cosΘ)

= 2secΘ2secΘ

Therefore, LHS = RHS

Hence proved

(iii) √(1+cosΘ)(1−cosΘ)(1+cosΘ)(1−cosΘ)−−−−−−√ \sqrt{\frac{(1-cos \Theta)}{(1+cos \Theta)}} = 2cosecΘ2cosecΘ

Ans:

To prove,

√(1−cosΘ)(1+cosΘ)+√(1+cosΘ)(1−cosΘ)(1−cosΘ)(1+cosΘ)−−−−−−√+(1+cosΘ)(1−cosΘ)−−−−−−√ = 2cosec \Theta

Considering left hand side (LHS),

Rationalize the numerator and denominator.

= √(1−cosΘ)(1−cosΘ)(1+cosΘ)(1−cosΘ)+√(1+cosΘ)(1+cosΘ)(1−cosΘ)(1+cosΘ)(1−cosΘ)(1−cosΘ)(1+cosΘ)(1−cosΘ)−−−−−−−−−−−−√+(1+cosΘ)(1+cosΘ)(1−cosΘ)(1+cosΘ)−−−−−−−−−−−−√

= √(1−cosΘ)2(1−cos2Θ)+√(1+cosΘ)2(1−cos2Θ)(1−cosΘ)2(1−cos2Θ)−−−−−−−√+(1+cosΘ)2(1−cos2Θ)−−−−−−−√

= √(1−cosΘ)2(sin2Θ)+√(1+cosΘ)2(sin2Θ)(1−cosΘ)2(sin2Θ)−−−−−−−√+(1+cosΘ)2(sin2Θ)−−−−−−−√

= (1−cosΘ)(sinΘ)+(1+cosΘ)(sinΘ)(1−cosΘ)(sinΘ)+(1+cosΘ)(sinΘ)

= (1−cosΘ+1+cosΘ)(sinΘ)(1−cosΘ+1+cosΘ)(sinΘ)

= (2)(sinΘ)(2)(sinΘ)

= 2cosec \Theta

Therefore, LHS = RHS

Hence proved

(iv) secΘ−1secΘ+1secΘ−1secΘ+1 = (sinΘ1+cosΘ)2(sinΘ1+cosΘ)2

Ans:

To prove,

secΘ−1secΘ+1secΘ−1secΘ+1 = (sinΘ1+cosΘ)2(sinΘ1+cosΘ)2

Considering left hand side (LHS),

= secΘ−1secΘ+1secΘ−1secΘ+1

= 1−cosΘ1+cosΘ1−cosΘ1+cosΘ

Multiply and divide with (1+cosΘΘ)

= (1−cosΘ)(1+cosΘ)(1+cosΘ)(1+cosΘ)(1−cosΘ)(1+cosΘ)(1+cosΘ)(1+cosΘ)

= (1−cos2Θ)(1+cosΘ)2(1−cos2Θ)(1+cosΘ)2

= (sin2Θ)(1+cosΘ)2(sin2Θ)(1+cosΘ)2

= (sinΘ1+cosΘ)2(sinΘ1+cosΘ)2

Therefore, LHS = RHS

Hence proved