Math, asked by MRDEMANDING, 2 months ago

prove that sintita -costita +1/ sintita tcostita
1-1/sectitattantita



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Answers

Answered by BrainlyEmpire
88

☯ To prove ☯

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

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☯ Given ☯⠀⠀

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Here, L.H.S. is \dfrac{sin\theta-cos\theta+1}{sin\theta+cos\theta-1}

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⚛ Step 1.

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In this step, we take the L.H.S. and divide its numerator and denominator by cos\theta\neq 0 in order to bring terms sec\theta and tan\theta of the R.H.S.

\Rightarrow \dfrac{tan\theta-1+sec\theta}{tan\theta+1-sec\theta}

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⚛ Step 2.

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In this step, we simply rearrange the terms of Step 1 in a suitable manner.

\Rightarrow \dfrac{tan\theta-(1-sec\theta)}{tan\theta+(1-sec\theta)}

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⚛ Step 3

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In this step, we multiply both the numerator and the denominator by tan\theta-(1-sec\theta).

\Rightarrow \dfrac{\{tan\theta-(1-sec\theta)\}^{2}}{\{tan\theta+(1-sec\theta)\}\{tan\theta-(1-sec\theta)\}}

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⚛ Step 4.

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Use the algebraic identities (a-b)^{2}=a^{2}-2ab+b^{2} and (a+b)(a-b)=a^{2}-b^{2}.

\Rightarrow \dfrac{tan^{2}\theta-2\:tan\theta\:(1-sec\theta)+(1-sec\theta)^{2}}{tan^{2}\theta-(1-sec\theta)^{2}}

\Rightarrow \dfrac{tan^{2}\theta-2\:tan\theta+2\:sec\theta\:tan\theta+1-2\:sec\theta+sec^{2}\theta}{tan^{2}\theta-1+2\:sec\theta-sec^{2}\theta}

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⚛ Step 5.

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In this step, we rearrange the terms in a manner so that we can use the trigonometric identity sec^{2}\theta-tan^{2}\theta=1.

\Rightarrow \dfrac{sec^{2}\theta+1+tan^{2}\theta+2\:sec\theta\:tan\theta-2\:sec\theta-2\:tan\theta}{-1-(sec^{2}\theta-tan^{2}\theta)+2\:sec\theta}

\Rightarrow \dfrac{sec^{2}\theta+sec^{2}\theta+2\:sec\theta\:tan\theta-2\:sec\theta-2\:tan\theta}{-1-1+2\:sec\theta}

\Rightarrow \dfrac{2\:sec^{2}\theta+2\:sec\theta\:tan\theta-2\:sec\theta-2\:tan\theta}{-2+2\:sec\theta}

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⚛ Step 6.

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In this step, we divide both numerator and denominator of Step 5 and try to factorize the terms with one factor sec\theta-1.

\Rightarrow \dfrac{sec^{2}\theta+sec\theta\:tan\theta-sec\theta-tan\theta}{-1+sec\theta}

\Rightarrow \dfrac{sec\theta\:(sec\theta+tan\theta)-1\:(sec\theta+tan\theta)}{sec\theta-1}

\Rightarrow \dfrac{(sec\theta+tan\theta)\:(sec\theta-1)}{sec\theta-1}

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⚛ Step 7.

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In this step, we divide both numerator and denominator by sec\theta-1\neq 0 and then we use trigonometric identity sec^{2}\theta-tan^{2}\theta=1 and algebraic identity a^{2}-b^{2}=(a+b)(a-b).

\Rightarrow sec\theta+tan\theta

\Rightarrow \dfrac{sec\theta+tan\theta}{1}

\Rightarrow \dfrac{sec\theta+tan\theta}{sec^{2}\theta-tan^{2}\theta}

\Rightarrow \dfrac{sec\theta+tan\theta}{(sec\theta+tan\theta)\:(sec\theta-tan\theta)}

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⚛ Step 8.

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To reach our the conclusion to our proof, we divide both numerator and denominator by sec\theta+tan\theta\neq 0.

\Rightarrow \dfrac{1}{sec\theta-tan\theta}

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⚛ Step 9. Conclusion step

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Thus, \dfrac{sin\theta-cos\theta+1}{sin\theta+cos\theta-1}=\dfrac{1}{sec\theta+tan\theta}

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• This completes our proof !

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Answered by dhairya348
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