Question

$\tt\texttt{If }m\,\cos\alpha - n\,\sin\alpha=p \texttt{, then prove that } \\\\ \tt m\,\sin\alpha+n\,\cos\alpha=\pm\sqrt{m^2+n^2-p^2}$

Subject: Mathematics
Topic: Trigonometry
Grade: 10th/11th

Answer 1

Answer:

Ur answer is in the attachment.

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Answer 2

Answer:

$\large \text{ (m \ sin\alpha +n \cos\alpha)= \pm\sqrt{m^2+n^2-p^2} Proved.}$

Step-by-step explanation:

Given :

$\large m \ cos\alpha -n \ sin\alpha=p \ .....( i )$

Now squaring on both side to ( i )

$\large (m \ cos\alpha -n \ sin\alpha)^2=(p)^2 \\\\\\\large \text{Using identity (a-b)^2=a^2+b^2-2ab we get}\\\\\\\large m^2 \ cos^2\alpha +n^2 \ sin^2-2mn \ sin\alpha \ cos\alpha=p^2 \ ...(ii)$

We know that

$\large cos^2\alpha +sin^2\alpha=1\\\\\\\large \text{We can also write as sin^2\alpha=1-cos^2\alpha \ and \ cos^2\alpha =1-sin^2\alpha }\\\\\\\large \text{Now putting these in ( ii )}$

$\large m^2 \ (1-sin^2\alpha) +n^2 \ (1-cos^2\alpha )-2mn \ sin\alpha \ cos\alpha=p^2 \\\\\\\large m^2 -m^2 \ sin^2\alpha+n^2-n^2 \ cos^2\alpha-2mn \ sin\alpha \ cos\alpha=p^2\\\\\\\large m^2 \ sin^2\alpha+n^2 \ cos^2\alpha+2mn \ sin\alpha \ cos\alpha=m^2+n^2-p^2\\\\\\\large (m \ sin\alpha +n \cos\alpha)^2= m^2+n^2-p^2\\\\\\\large (m \ sin\alpha +n \cos\alpha)= \pm\sqrt{m^2+n^2-p^2}$

L.H.S. = R.H.S.

Hence proved.