Math, asked by rahulsihag33, 1 year ago

Please solve above the QUESTION

Attachments:

shadowsabers03: Hey, bro. What's the continuation of the question? It's not appeared.

shadowsabers03: We have to prove it using something. That's the question. Please say what is to be used.

Answers

Answered by shadowsabers03

Hey mate!

The question is wrong due to some technical error. The actual question is given below.

Question:

$$$Prove that$ \\ \\ \\ \boxed{\frac{\sin\theta+\cos\theta-1}{\sin\theta-\cos\theta+1}=\frac{1}{\bold{sec \theta}+\tan\theta}}$

Proof:

We know that,

$\sec^2\theta-\tan^2\theta=1 \\ \\ \\ \therefore\ -(\sec^2\theta-\tan^2\theta)=-1 \\ \\ \Rightarrow\ \tan^2\theta-\sec^2\theta=-1$

$\ \ \ \ \ \boxed{LHS} \\ \\ \\ \Rightarrow\ \boxed{\frac{\sin\theta+\cos\theta-1}{\sin\theta-\cos\theta+1}} \\ \\ \\ \Rightarrow\ \boxed{\frac{(\sin\theta+\cos\theta-1)\div\cos\theta}{(\sin\theta-\cos\theta+1)\div\cos\theta}} \\ \\ \\ \Rightarrow\ \boxed{\frac{\frac{\sin\theta+\cos\theta-1}{\cos\theta}}{\frac{\sin\theta-\cos\theta+1}{\cos\theta}}}$

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

$\Rightarrow\ \boxed{\frac{(\tan\theta-\sec\theta+1)(\tan\theta+\sec\theta)}{(\tan\theta+\sec\theta-1)(\tan\theta+\sec\theta)}} \\ \\ \\ \Rightarrow\ \boxed{\frac{(\tan\theta-\sec\theta)(\tan\theta+\sec\theta)+1(\tan\theta+\sec\theta)}{(\tan\theta+\sec\theta-1)(\tan\theta+\sec\theta)}} \\ \\ \\ \Rightarrow\ \boxed{\frac{\tan^2\theta-\sec^2\theta+\tan\theta+\sec\theta}{(\tan\theta+\sec\theta-1)(\tan\theta+\sec\theta)}}$

$\Rightarrow\ \boxed{\frac{-1+\tan\theta+\sec\theta}{(\tan\theta+\sec\theta-1)(\tan\theta+\sec\theta)}} \\ \\ \\ \Rightarrow\ \boxed{\frac{\tan\theta+\sec\theta-1}{(\tan\theta+\sec\theta-1)(\tan\theta+\sec\theta)}} \\ \\ \\ \Rightarrow\ \boxed{\frac{1}{\tan\theta+\sec\theta}} \\ \\ \\ \Rightarrow\ \boxed{\frac{1}{\sec\theta+\tan\theta}} \\ \\ \\ \Rightarrow\ \boxed{RHS}$

Hence proved!

Hope this helps. Please mark it as the brainliest.

Please ask me if you've any doubts.

Thank you. :-))

rahulsihag33: hii bro this is not a right answer

shadowsabers03: Bro., the question was wrong. It would never be sin.

shadowsabers03: It's sec. Sometimes it was due to technical error in printing.

shadowsabers03: We can't prove your question. It can never be.

shadowsabers03: But, I have to ask you one thing. The question tells to use one identity, but it can't be seen in the attachment. What identity is to be used?

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