Question

prove this trigonometric equation​

Answer 1

⇒ sinθtanθ/(1 - cosθ)

⇒ sinθ(sinθ/cosθ) / (1 - cosθ)

                   [ 1/cosA = secA ]

⇒ sinθ(sinθsecθ) / (1 - cosθ)

⇒ sin²θsecθ / (1 - cosθ)

                   [ sin²A = 1 - cos²A ]

⇒ (1 - cos²θ)secθ /(1 - cosθ)

⇒ (1 - cosθ)(1 + cosθ)secθ/(1 - cosθ)

⇒ (1 + cosθ)secθ

⇒ secθ + cosθ.secθ

⇒ secθ + 1              {cosθ.secθ = 1}

Answer 2

$\frac{ \sin \theta \tan\theta}{1 - \cos\theta } = 1 + \sec\theta \\ \\ \\ \text{Solving the LHS } \\ \\ \implies \tt \frac{ \sin\theta \bigg \{ \dfrac{ \sin\theta }{ \cos\theta } \bigg \}}{1 - \cos\theta } \\ \\ \implies \frac{ \sin\theta \sin\theta }{ \cos\theta \{1 - \cos(\theta) \}} \\ \\ \\ \implies \tt \frac{ { \sin }^{2}\theta }{ \cos \theta\{ 1 - \cos\theta \} } \\ \\ \\ \implies \tt \frac{1 - { \cos}^{2} \theta}{\cos \theta \{ 1 - \cos\theta \}} \\ \\ \\ \implies \tt \frac{(1 + \cos\theta)(1 - \cos\theta)}{\cos \theta (1 - \cos\theta )} \\ \\ \\ \implies \tt \frac{(1 + \cos\theta) \cancel{(1 - \cos\theta)}}{\cos \theta \cancel{ (1 - \cos\theta )}} \\ \\ \\ \implies \tt \frac{1 + \cos\theta}{\cos\theta} \\ \\ \implies \tt \frac{1}{ \cos\theta} + \frac{\cos\theta}{\cos\theta} \\ \\ \implies \tt \sec\theta + 1 \\ \implies \sf 1 + \sec\theta$

Thus,

LHS = RHS

Used Ratios & identities of trigonometry :

• tanθ = sinθ/cosθ

• sin²θ + cos²θ = 1

→ sin²θ = 1 - cos²θ

• cosθ = 1/secθ

• sec = 1/cosθ

Other important ratios of trigonometry :

• sinθ = 1/cosecθ

• cotθ = 1/tanθ = cosθ/sinθ

• sec²θ - tan²θ = 1
• cosec²θ - cot²θ = 1

• sec²θ = 1 + tan²θ
• cosec²θ = 1 + cot²θ

• sin²θ = 1 - cos²θ