Question

plz solve this problem of trigonometry​

Answer 1

Solution :

$\bf L.H.S = \tt \dfrac{sec\: \theta + tan \: \theta - 1}{tan \: \theta - sec \: \theta + 1}$

$: \implies \tt \dfrac{\frac{1}{cos \: \theta} + \frac{sin \: \theta}{cos \: \theta} - 1}{ \frac{sin \: \theta}{cos \: \theta} - \frac{1}{cos \: \theta} + 1 } \: = \dfrac{1 + sin \: \theta - cos \: \theta}{sin \: \theta + cos \: \theta}$

$: \implies \tt\dfrac{ sin \: \theta - (cos \: \theta - 1)}{sin \: \theta + (cos \: \theta - 1)} \: \times \: \dfrac{ sin \: \theta - (cos \: \theta - 1)}{sin \: \theta - (cos \: \theta - 1)}$

$: \implies \tt\dfrac{ sin^{2} \: \theta + cos^{2} \: \theta + 1 - 2 \: cos \: \theta - 2 \: sin \: \theta \: (cos \: \theta - 1)}{sin^{2} \: \theta - (cos \: \theta - 1)^{2} }$

$: \implies \tt\dfrac{1 + 1 - 2 \: cos \: \theta - 2 \: sin \: \theta \: cos \: \theta + 2 \: sin \: \theta}{sin^{2} \: \theta + cos^{2} \: \theta - 1 + 2 \: cos \: \theta }$

$: \implies \tt\dfrac{2 - 2 \: cos \: \theta - 2 \: sin \: \theta \: cos \: \theta + 2 \: sin \: \theta}{sin^{2} \: \theta + cos^{2} \: \theta - sin^{2} \: \theta - cos^{2} \: \theta + 2 \: cos \: \theta }$

$: \implies \tt\dfrac{2 (1 - \: cos \: \theta )- 2 \: sin \: \theta (1 - \: cos \: \theta)}{ 2 \: cos \: \theta - 2 \: cos^{2} \: \theta}$

$: \implies \tt\dfrac{(2 + 2 \: sin \: \theta) \: \cancel{(1 - cos\: \theta)}}{2 \: cos \: \theta \: \cancel{(1 - cos \: \theta)}} \: = \: \dfrac{1 + sin \: \theta}{cos \: \theta}$

$: \implies\tt\dfrac{1 + sin \: \theta}{cos \: \theta} \: \times \: \dfrac{1 - sin \: \theta}{1 - sin \: \theta}$

$: \implies\tt\dfrac{1 + sin^{2} \: \theta}{cos \: (1 - sin \: \theta)}$

$: \implies\tt\dfrac{cos^{2} \: \theta}{cos \: \theta (1 - sin \: \theta)}$

$: \implies\tt\dfrac{cos \: \theta}{1 - sin \: \theta} \: = \: \bf{ R.H.S}$

$\huge\bigstar \:\underline{\red{\sf Hence, Proved}} \: \bigstar$

Answer 2

To prove :

$\purple{\bigstar}\;\;\dfrac{sec\;\theta+tan\;\theta-1}{tan\;\theta-sec\;\theta+1}=\dfrac{cos\;\theta}{1-sin\;\theta}$

Proof :

Taking LHS

$\longrightarrow LHS = \dfrac{sec\;\theta+tan\;\theta-1}{tan\;\theta-sec\;\theta+1}$

using trigonometric identity

sec²θ - tan²θ = 1

$\longrightarrow LHS = \dfrac{sec\;\theta+tan\;\theta-(sec^2\theta-tan^2\theta)}{tan\;\theta-sec\;\theta+1}$

Using algebraic identity

a² - b² = ( a + b ) ( a - b )

$\longrightarrow LHS = \dfrac{sec\;\theta+tan\;\theta-(sec\;\theta-tan\;\theta)(sec\;\theta+tan\;\theta)}{tan\;\theta-sec\;\theta+1}$

Taking sec θ + tan θ common in numerator

$\longrightarrow LHS = \dfrac{(sec\;\theta+tan\;\theta)(1-(sec\;\theta-tan\;\theta))}{tan\;\theta-sec\;\theta+1}$

$\longrightarrow LHS = \dfrac{(sec\;\theta+tan\;\theta)(1-sec\;\theta+tan\;\theta)}{tan\;\theta-sec\;\theta+1}$

$\longrightarrow LHS = \dfrac{(sec\;\theta+tan\;\theta)(tan\;\theta-sec\;\theta+1)}{tan\;\theta-sec\;\theta+1}$

$\longrightarrow LHS = sec\;\theta + tan\;\theta$

putting sec θ = 1 /cos θ and tan θ = sin θ / cos θ

$\longrightarrow LHS = \dfrac{1}{cos\;\theta}+\dfrac{sin\;\theta}{cos\;\theta}$

Taking LCM

$\longrightarrow LHS = \dfrac{1+sin\;\theta}{cos\;\theta}$

Multiplying by cos θ in numerator and denominator

$\longrightarrow LHS = \dfrac{1+sin\;\theta}{cos\;\theta}\times \dfrac{cos\;\theta}{cos\;\theta}$

$\longrightarrow LHS = \dfrac{1+sin\;\theta(cos\;\theta)}{cos^2\theta}$

Using trigonometric identity

1 - sin²θ = cos²θ

$\longrightarrow LHS = \dfrac{1+sin\;\theta(cos\;\theta)}{1-sin^2\theta}$

Using algebraic identity

a² - b² = ( a + b ) ( a - b )

$\longrightarrow LHS = \dfrac{(1+sin\;\theta)(cos\;\theta)}{(1+sin\;\theta)(1-sin\;\theta)}$

$\longrightarrow LHS = \dfrac{cos\;\theta}{1-sin\;\theta} = RHS$

PROVED .