Prove that:
cot 90°
)
cot 30° cot 60° - 1
cot 30° + cot 60°
Answers
Answer:
\dfrac{\cot 30.\cot 60-1}{\cot 30+\cot 60}=\cot 90
cot30+cot60
cot30.cot60−1
=cot90 , proved.
Step-by-step explanation:
Prove that, \dfrac{\cot 30.\cot 60-1}{\cot 30+\cot 60}=\cot
cot30+cot60
cot30.cot60−1
=cot 90[/tex]
L.H.S. = \dfrac{\cot 30.\cot 60-1}{\cot 30+\cot 60}
cot30+cot60
cot30.cot60−1
=\dfrac{\sqrt{3} \times \frac{1}{\sqrt{3}} -1}{\sqrt{3} +\frac{1}{\sqrt{3}}}
3
+
3
1
3
×
3
1
−1
[ ∵ \cot 30=\sqrt{3}cot30=
3
and \cot 60=\dfrac{1}{\sqrt{3}}cot60=
3
1
=\dfrac{1 -1}{\frac{3+1}{\sqrt{3}}}=
3
3+1
1−1
=\dfrac{0}{\frac{3+1}{\sqrt{3}}}=
3
3+1
0
= 0
R.H.S. = \cot 90cot90
= 0
[ ∵ \cot 90=0cot90=0
L.H.S. = R.H.S. = 0, proved.
Step-by-step explanation:
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