Answer
Verified
410.4k+ views
Hint- Start with taking out cot$3x$ common from the last two terms and then write $3x$ as $2x+x$ , now use the identity \[\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}\] .
Complete step-by-step answer:
Given equation is $\cot x.\cot 2x - \cot 2x.\cot 3x - \cot x.\cot 3x = 1$
By taking L.H.S we have
$ = \cot x.\cot 2x - \cot 2x.\cot 3x - \cot x.\cot 3x$
Take $\cot 3x$ common from above term.
$\cot x.\cot 2x - \cot 3x\left[ {\cot 2x + \cot x} \right]$
Here we can write \[\cot 3x{\text{ as }}\cot \left( {x + 2x} \right)\]
Therefore the term becomes
$ = \cot x.\cot 2x - \cot \left( {x + 2x} \right)\left[ {\cot 2x + \cot x} \right]$
AS we know that $\left[ {\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}} \right]$
By applying this formula, we get
\[ \Rightarrow \cot x\cot 2x - \left( {\dfrac{{\cot 2x\cot x - 1}}{{\cot 2x + \cot x}}} \right)\left( {\cot 2x + \cot x} \right)\]
Now by simplifying the above term, we get
$
\Rightarrow \cot 2x\cot x - \left( {\cot 2x\cot x - 1} \right) \\
\Rightarrow \cot 2x\cot x - \cot 2x\cot x + 1 \\
\Rightarrow 1 \\
$
Hence, the result is same as R.H.S
Note- In order to solve these types of problems, first of all remember all the trigonometric identities and formulas. In the above question we use a formula of cotangent to split its angles. Similarly remember formulas to change product into sum and sum into product.
Complete step-by-step answer:
Given equation is $\cot x.\cot 2x - \cot 2x.\cot 3x - \cot x.\cot 3x = 1$
By taking L.H.S we have
$ = \cot x.\cot 2x - \cot 2x.\cot 3x - \cot x.\cot 3x$
Take $\cot 3x$ common from above term.
$\cot x.\cot 2x - \cot 3x\left[ {\cot 2x + \cot x} \right]$
Here we can write \[\cot 3x{\text{ as }}\cot \left( {x + 2x} \right)\]
Therefore the term becomes
$ = \cot x.\cot 2x - \cot \left( {x + 2x} \right)\left[ {\cot 2x + \cot x} \right]$
AS we know that $\left[ {\cot \left( {A + B} \right) = \dfrac{{\cot A\cot B - 1}}{{\cot A + \cot B}}} \right]$
By applying this formula, we get
\[ \Rightarrow \cot x\cot 2x - \left( {\dfrac{{\cot 2x\cot x - 1}}{{\cot 2x + \cot x}}} \right)\left( {\cot 2x + \cot x} \right)\]
Now by simplifying the above term, we get
$
\Rightarrow \cot 2x\cot x - \left( {\cot 2x\cot x - 1} \right) \\
\Rightarrow \cot 2x\cot x - \cot 2x\cot x + 1 \\
\Rightarrow 1 \\
$
Hence, the result is same as R.H.S
Note- In order to solve these types of problems, first of all remember all the trigonometric identities and formulas. In the above question we use a formula of cotangent to split its angles. Similarly remember formulas to change product into sum and sum into product.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE