Question

(Prove that)
cos 14°– sin 14°
cos 140+ sin 14°
= cot 59°
plz help me to do this.
if anyone solve this for me then i will mark him/her brilliant.​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Correct Question

Prove that    

$\frac{cos14\textdegree-sin 14\textdegree}{cos14\textdegree+sin 14\textdegree} = cot \ 59\textdegree$

Answer:

It is proved that    $\frac{cos14\textdegree-sin 14\textdegree}{cos14\textdegree+sin 14\textdegree} = cot \ 59\textdegree$.

Step-by-step explanation:

Given:

$\frac{cos14\textdegree-sin 14\textdegree}{cos14\textdegree+sin 14\textdegree} = cot \ 59\textdegree$

To Find:

It is to be proved that $\frac{cos14\textdegree-sin 14\textdegree}{cos14\textdegree+sin 14\textdegree} = cot \ 59\textdegree$.

Formula Used:

$tan(A-B)=\frac{tanA-tanB}{1+tanA.tanB}$

Solution:

As given- $\frac{cos14\textdegree-sin 14\textdegree}{cos14\textdegree+sin 14\textdegree} = cot \ 59\textdegree$

 $LHS=\frac{cos14\textdegree-sin 14\textdegree}{cos14\textdegree+sin 14\textdegree}$

Divide the numerator and denominator by $cos14 \textdegree$.

$LHS=\frac{\frac{cos14\textdegree}{cos14\textdegree} -\frac{sin14\textdegree}{cos14\textdegree} }{\frac{cos14\textdegree}{sin14\textdegree} +\frac{sin14\textdegree}{cos14\textdegree} }$

        $=\frac{1-tan14\textdegree}{1+tan14\textdegree}$

        $=\frac{1-tan14\textdegree}{1+ 1.tan14\textdegree}$

Putting $1=tan45\textdegree$.

       $=\frac{tan45\textdegree-tan14\textdegree}{1+tan45\textdegree.tan14\textdegree}$      

       $= tan( 45\textdegree-14\textdegree)$  

       $= tan31\textdegree$

       $= tan(90\textdegree-59\textdegree)$

       $=cot 59\textdegree$

       $=RHS$

$LHS=RHS$

Hence, it is It is proved that    $\frac{cos14\textdegree-sin 14\textdegree}{cos14\textdegree+sin 14\textdegree} = cot \ 59\textdegree$.