Question

Prove the following trigonometric identities:
(cosecθ-secθ)(cotθ-tanθ)=(cosecθ+secθ)(secθcosecθ-2)

Answer 1

The equality is $(cosec\theta-sec\theta)(cot\theta-tan\theta)=(cosec\theta+sec\theta)(sec\theta cosec\theta-2)$

is proved

Step-by-step explanation:

• To prove that $(cosec\theta-sec\theta)(cot\theta-tan\theta)=(cosec\theta+sec\theta)(sec\theta cosec\theta-2)$
• That is prove that LHS=RHS
• Let take LHS

• $(cosec\theta-sec\theta)(cot\theta-tan\theta)$
• $(\frac{1}{sin\theta}-\frac{1}{cos\theta})(\frac{cos\theta}{sin\theta}-\frac{sin\theta}{cos\theta})$
• $(\frac{1(cos\theta)-1(sin\theta)}{sin\theta cos\theta})(\frac{cos\theta(cos\theta)-sin\theta(sin\theta)}{sin\theta cos\theta})$
• $(\frac{cos\theta-sin\theta}{sin\theta cos\theta})(\frac{cos^2\theta-sin^2\theta}{sin\theta cos\theta})$
• $(\frac{cos\theta-sin\theta}{sin\theta cos\theta})(\frac{(cos\theta-sin\theta)(cos\theta+sin\theta)}{sin\theta cos\theta})$

• $(\frac{cos\theta-sin\theta)((cos\theta-sin\theta)(cos\theta+sin\theta))}{sin^2\theta cos^2\theta}$ ( by using the identity $(a^2-b^2)=(a+b)(a-b)$ )

$\frac{(cos\theta-sin\theta)^2(cos\theta+sin\theta)}{sin^2\theta cos^2\theta}$=LHS

• Now RHS $(cosec\theta+sec\theta)(sec\theta cosec\theta-2)$
• $=\frac{1}{sin\theta}+\frac{1}{cos\theta})(\frac{1}{cos\theta}\frac{1}{sin\theta}-2)$
• $=(\frac{cos\theta+sin\theta}{sin\theta cos\theta})(\frac{1-2sin\theta cos\theta}{cos\theta sin\theta})$

• $=(\frac{cos\theta+sin\theta}{sin\theta cos\theta})(\frac{cos^2\theta+sin^2\theta-2sin\theta cos\theta}{cos\theta sin\theta})$( by using the identity $cos^2x+sin^2x=1$ )

• $=(\frac{cos\theta+sin\theta}{sin\theta cos\theta})(\frac{(cos\theta+sin\theta)^2}{cos\theta sin\theta})$    ( by using the identity $(a+b)^2=a^2+2ab+b^2$ )

$=\frac{(cos\theta-sin\theta)^2(cos\theta+sin\theta)}{sin^2\theta cos^2\theta}$=RHS

Comparing LHS and RHS we get

LHS=RHS

Therefore $(cosec\theta-sec\theta)(cot\theta-tan\theta)=(cosec\theta+sec\theta)(sec\theta cosec\theta-2)$