Can any one solve this no. 48
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Solution:
Given to prove:
[tex] \tt \longrightarrow \dfrac{cos \: \theta \: cot \: \theta}{1 + sin \: \theta} =cosec \: \theta - 1[/tex]
Taking Left Hand Side, we get:
[tex] \tt = \dfrac{cos \: \theta \: cot \: \theta}{1 + sin \: \theta}[/tex]
Multiplying both numerator and denominator by (1 - sin θ), we get:
[tex] \tt = \dfrac{cos \: \theta \: cot \: \theta \: (1 - sin\: \theta)}{(1 + sin \: \theta)(1 - sin\: \theta)}[/tex]
[tex] \tt = \dfrac{cos \: \theta \: cot \: \theta \: (1 - sin\: \theta)}{1 - {sin}^{2} \theta}[/tex]
Substituting cos²θ = 1 - sin²θ, we get:
[tex] \tt = \dfrac{cos \: \theta \:cot \: \theta \: (1 - sin\: \theta)}{{cos}^{2} \theta}[/tex]
[tex] \tt = \dfrac{cot \: \theta \: (1 - sin\: \theta)}{cos \: \theta}[/tex]
Substituting cot θ = cos θ / sin θ, we get:
[tex] \tt = \dfrac{ \dfrac{cos \: \theta}{sin \: \theta} (1 - sin\: \theta)}{cos \: \theta}[/tex]
[tex] \tt = \dfrac{cos \: \theta \: (1 - sin\: \theta)}{cos \: \theta \: sin \: \theta}[/tex]
[tex] \tt = \dfrac{1 - sin\: \theta}{sin \: \theta}[/tex]
[tex] \tt = \dfrac{1}{sin \: \theta} - \dfrac{sin \:\theta}{sin \: \theta} [/tex]
[tex] \tt =cosec \: \theta - 1[/tex]
[tex] \tt = Right\:Hand\:Side[/tex]
Therefore:
[tex] \tt \longrightarrow \dfrac{cos \: \theta \: cot \: \theta}{1 + sin \: \theta} =cosec \: \theta - 1[/tex]
Hence Proved..!!
Learn More:
1. Relationship between sides and T-Ratios.
2. Square formulae.
3. Reciprocal Relationship.
4. Cofunction identities.
5. Even odd identities.