Question

Hey any maths aryabhatta , plz solve this question ...... ✌​

Answer 1

Question -

Given

If 2x and 2y are complementary

And tan ( x + 2y ) = 2

To find :-

The value of cot ( x - y ) .

Solution :-

We are given that 2x and 2y are complementary angle .

Sum of complementary angles is equal to 90° .

↭ 2x + 2y = 90 °

From here we'll get the value of 2y .

↭ 2y = 90 - 2x.

Now x + y = $\frac{90}{2}$

↭ x + y = 45°

From here we'll find out the value of x .

↭ x = 45 - y .

Now we are given that tan( x + 2y ) = 2 . So ,

Putting the value of 2y here ie 2y = 90-2x

↭ tan ( x + 2y ) = 2

↭ tan ( x + 90 - 2x ) = 2

↭ tan ( 90 - x) = 2

As we know that tan(90-A) = Cot A .

↭ Cot X = 2

We got the value of cot X ie 2

Now putting the value of x in the given equation .

↭ tan ( x + 2y ) = 2

↭ tan ( 45 - y + 2y ) = 2

↭ tan ( 45 + y ) = 2

Now using tan ( 45 + A ) expansion .

We know that tan (45 + y) = $\frac{ \tan 45 + \tan y}{ 1 - \tan 45 . \tan y }$

As we know that tan 45° = 1 .So

$\frac{1 + \tan y}{1 - \tan y }$ = $\frac{2}{1}$

Now applying componendo and dividendo

1 + tan y + 1 - tan y / 1 + tan y - 1 + tan y = 2 +1/2-1

2 / 2 tan y = 3 / 1

1 / tan y = 3

So cot y = 3 .

We got the value of cot y also .

Now expanding cot ( x - y ) by sum and difference of angles .

$\frac{\cot (x) \times \cot (y) + 1 }{ \cot (y) - \cot (x)}$

Now putting the value of cot X and cot y over here.

↭ ( 2 × 3 ) + 1 / 3-2

↭ 6 + 1 / 1

↭ 7

Cot ( x - y ) = 7 .

Answer 2

Question :

If 2x and 2y are complementary angles and tan(x+2y)=2, then find the value of cot(x-2y)

Formula's used :

•Trignometric Formula's:

$\sf \tan(x + y) = \frac{ \tan x + \tan y}{1 - \tan x \tan y}$

$\sf \cot(x - y) = \frac{ \cot x \cot y + 1}{ \cot y - \cot x}$

•Componendo and dividendo:

For any equal ratio

$\dfrac{a}{b} = \dfrac{c}{d}$

we can apply Componendo and dividendo, which gives

$\dfrac{a + b}{a - b} = \dfrac{c + d}{c - d}$

Solution :

Given : 2x and 2y are complementary angles .

⇒ 2x+2y = 90°

⇒x+y=45°

⇒y=45°-x

$\sf \tan(x + 2y) = 2$

Put y = 45-x

$\implies \sf \tan(x + 2(45 - x)) = 2$

$\implies \sf \tan(90 - x) = 2$

we know that tan (90-x)= cotx

$\implies \sf \cot x = 2 ...(1)$

And $\tan(x + 2y) = 2$

put x = 45-y

$\implies \sf \tan(45 - y + 2y) = 2$

$\implies \sf \tan(45 + y) = 2$

$\implies \sf \frac{ \tan45 + \tan y}{1 - \tan45 \tan y} = 2$

$\implies \sf \frac{1 + \tan y}{1 - \tan y} = 2$

Now use Componendo and dividendo

$\implies \sf \frac{(1 + \tan y) + (1 - \tan y)}{(1 + \tan y) - (1 - \tan y)} = \frac{2 + 1}{2 - 1}$

$\implies \sf \frac{ 1 + \tan y + 1 - \tan y}{1 + \tan y - 1 + \tan y} = \frac{3}{1}$

$\implies \sf \dfrac{ \cancel{2}}{ \cancel{2} \times \tan y} = 3$

$\implies \sf \cot y = 3 ...(2)$

Now cot (x-y)

$= \sf \dfrac{ \cot x \cot y - 1}{ \cot y - \cot x}$

put the values of cot x and cot y from (1)and (2)

$= \sf \dfrac{2 \times 3 + 1}{3 - 2}$

$= 7$

Therefore,the value of cot(x-y)=7

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More Trigonometric Formula's:

sin2A = 2 sinA cosA

cos2A = cos²A - sin²A

tan2A = 2 tanA / (1 - tan²A)