Math, asked by arshdeepsingh4551, 3 months ago

prove that Maths sum 10th class trignometry 8th chapter ​

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Answered by akuldas19
4

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Answered by Asterinn
3

Given :

 \rm \dfrac{tan A-Sin A }{tan A + Sin A } =  \dfrac{Sec A-1}{Sec A + 1}

To prove :

LHS = RHS

Proof :

\rm RHS =  \dfrac{Sec A-1}{Sec A + 1}

\rm \:LHS =    \dfrac{tan A-Sin A }{tan A + Sin A }

 \rm  \rightarrow\dfrac{tan A-Sin A }{tan A + Sin A }

 \rm  \rightarrow\dfrac{ \dfrac{Sin  \: A}{cos \: A} -\dfrac{Sin  \: A}{ 1} }{ \dfrac{Sin  \: A}{cos \: A}  + \dfrac{Sin  \: A}{ 1}}  \\  \\ \rm  \rightarrow \:  \dfrac{Sin  \: A}{Sin  \: A}  \times  \bigg(\dfrac{ \dfrac{1}{cos \: A} -1 }{ \dfrac{1}{cos \: A}  +1}\bigg ) \\  \\ \rm  \rightarrow \:  1 \times \dfrac{ secA -1 }{  secA +1}\\  \\ \rm  \rightarrow \:  \dfrac{ secA -1 }{  secA +1}

Therefore, LHS = RHS

hence proved

Additional Information :-

\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ $\: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1$\end{minipage}}

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