Question

prove that sin theta minus cos theta + 1 by sin theta + cos theta minus one is equal to one by sec theta minus 10 theta using the identity sec squared theta is equal to one plus tan square theta

Answer 1

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Hey there !!


→ Prove that :-)


$\huge \bf{ \frac{ sin \theta - cos \theta + 1 }{ sin \theta + cos \theta - 1 } = \frac{1}{sec \theta - tan \theta } }$


→ Solution :-)


$\boxed{ See \: the \: attachment.}$



$\boxed { \large\mathbb{TRIGONOMETRY.}}$



$\huge \bf \underline{ \mathbb{LHS = RHS.}}$




✔✔Hence, it is proved ✅✅.

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$\huge \boxed{ \mathcal{THANKS}}$




$\huge \bf{ \# \mathbb{B}e \mathbb{B}rainly.}$

Answer 2

Answer:

Step-by-step explanation:

Problem

$\frac{sin\theta-cos\theta+1}{sin\theta+cos\theta-1} =\frac{1}{sec\theta-tan\theta}$

Prove that RHS=LHS

RHS

$\frac{sin\theta-cos\theta+1}{sin\theta+cos\theta-1}\\\\=\frac{\frac{sin\theta-cos\theta+1}{cos\theta}}{\frac{sin\theta+cos\theta-1}{cos\theta}}\\\\=\frac{\frac{sin\theta}{cos\theta}-1+\frac{1}{cos\theta}}{\frac{sin\theta}{cos\theta}+1-\frac{1}{cos\theta}}\\\\=\frac{tan\theta-1+sec\theta}{tan\theta+1-sec\theta}\\\\=\frac{sec\theta+tan\theta-1}{1-(sec\theta-tan\theta)}\\\\=\frac{sec\theta+tan\theta-1}{(sec^{2}\theta-tan^{2}\theta)-(sec\theta-tan\theta)}\text{ [we know that }sec^{2}\theta-tan^{2}\theta=1]$

$\\\\=\frac{sec\theta+tan\theta-1}{(sec\theta+tan\theta)(sec\theta-tan\theta)-(sec\theta-tan\theta)}\\\\=\frac{sec\theta+tan\theta-1}{(sec\theta-tan\theta)(sec\theta+tan\theta-1)}=\frac{1}{(sec\theta-tan\theta)}$

RHS=LHS (proved)