Question

Sec teta + tan teta - 1/ tan teta - sec teta +1 = cod teta / 1- sin teta



Pls answer fast who ever is answering I will mark them as brainlist

I will cry pls give me answer

Answer 1

Solution:

See in the attachment.

Some Important Trigonometry Rules:

sinø . cosecø = 1

cosø . secø = 1

tanø . cotø = 1

sin²ø + cos²ø = 1

cosec²ø - cot² = 1

sec²ø - tan²ø = 1

Answer 2

Taking $\sf{\sec\theta=\dfrac{1}{\cos\theta}}$ and $\sf{\tan\theta=\dfrac{\sin\theta}{\cos\theta},}$ LHS becomes,

$\longrightarrow\sf{\dfrac{\sec\theta+\tan\theta-1}{\tan\theta-\sec\theta+1}=\dfrac{\left(\dfrac{1}{\cos\theta}+\dfrac{\sin\theta}{\cos\theta}-1\right)}{\left(\dfrac{\sin\theta}{\cos\theta}-\dfrac{1}{\cos\theta}+1\right)}}$

Taking $\sf{1}$ as $\sf{\dfrac{\cos\theta}{\cos\theta},}$

$\longrightarrow\sf{\dfrac{\sec\theta+\tan\theta-1}{\tan\theta-\sec\theta+1}=\dfrac{\left(\dfrac{1}{\cos\theta}+\dfrac{\sin\theta}{\cos\theta}-\dfrac{\cos\theta}{\cos\theta}\right)}{\left(\dfrac{\sin\theta}{\cos\theta}-\dfrac{1}{\cos\theta}+\dfrac{\cos\theta}{\cos\theta}\right)}}$

$\longrightarrow\sf{\dfrac{\sec\theta+\tan\theta-1}{\tan\theta-\sec\theta+1}=\dfrac{\left(\dfrac{1+\sin\theta-\cos\theta}{\cos\theta}\right)}{\left(\dfrac{\sin\theta-1+\cos\theta}{\cos\theta}\right)}}$

Since $\sf{\dfrac{\left(\dfrac{a}{c}\right)}{\left(\dfrac{b}{c}\right)}}$ can be taken as $\sf{\dfrac{a}{b},}$

$\longrightarrow\sf{\dfrac{\sec\theta+\tan\theta-1}{\tan\theta-\sec\theta+1}=\dfrac{1+\sin\theta-\cos\theta}{\sin\theta-1+\cos\theta}}$

$\longrightarrow\sf{\dfrac{\sec\theta+\tan\theta-1}{\tan\theta-\sec\theta+1}=\dfrac{\sin\theta+1-\cos\theta}{\sin\theta-1+\cos\theta}}$

Multiplying $\sf{1-\sin\theta}$ on both numerator and denominator in RHS,

$\longrightarrow\sf{\dfrac{\sec\theta+\tan\theta-1}{\tan\theta-\sec\theta+1}=\dfrac{(\sin\theta+1-\cos\theta)(1-\sin\theta)}{(\sin\theta-1+\cos\theta)(1-\sin\theta)}}$

$\longrightarrow\sf{\dfrac{\sec\theta+\tan\theta-1}{\tan\theta-\sec\theta+1}=\dfrac{(1+\sin\theta)(1-\sin\theta)-\cos\theta(1-\sin\theta)}{(\sin\theta-1+\cos\theta)(1-\sin\theta)}}$

$\longrightarrow\sf{\dfrac{\sec\theta+\tan\theta-1}{\tan\theta-\sec\theta+1}=\dfrac{1-\sin^2\theta-\cos\theta(1-\sin\theta)}{(\sin\theta-1+\cos\theta)(1-\sin\theta)}}$

Since $\sf{1-\sin^2\theta=\cos^2\theta,}$

$\longrightarrow\sf{\dfrac{\sec\theta+\tan\theta-1}{\tan\theta-\sec\theta+1}=\dfrac{\cos^2\theta-\cos\theta(1-\sin\theta)}{(\sin\theta-1+\cos\theta)(1-\sin\theta)}}$

$\longrightarrow\sf{\dfrac{\sec\theta+\tan\theta-1}{\tan\theta-\sec\theta+1}=\dfrac{\cos\theta(\sin\theta-1+\cos\theta)}{(\sin\theta-1+\cos\theta)(1-\sin\theta)}}$

$\longrightarrow\underline{\underline{\sf{\dfrac{\sec\theta+\tan\theta-1}{\tan\theta-\sec\theta+1}=\dfrac{\cos\theta}{1-\sin\theta}}}}$

Hence Proved!