Inverse Trignometric Functions

Solve the following equations:

Question 13: `2text(tan)^(-1)(text(cos) x)=text(tan)^(-1)(2text(cosec) x)`

Solution: Given, `2text(tan)^(-1)(text(cos) x)=text(tan)^(-1)(2text(cosec) x)`

Or, `text(tan)^(-1)((2text(cos) x)/(1-text(cos)^2x)``=text(tan)^(-1)((2x)/(1-x^2))`

Because `2text(tan)^(-1)x=text(tan)^(-1)(2x)/(1-x^2)`

Or, `text(tan)^(-1)((2text(cos) x)/(text(sin)^2x))=text(tan)^(-1)(2/(text(sin) x))`

Or, `(2text(cos) x)/(text(sin)^2x)=2/(text(sin) x)`

Or, `(2text(cos) x)/(text(sin)^2x)xx(text(sin) x)/2=1`

Or, `(text(cos) x)/(text(sin) x)=1`

Or, `text(cot) x=1`

Or, `x=text(cot)^(-1)(1)`

Or, `x=(π)/4`

Question 14: `text(tan)^(-1)(1-x)/(1+x)=1/2text(tan)^(-1)x`, (`x>0`)

Solution: Given, `text(tan)^(-1)(1-x)/(1+x)=1/2text(tan)^(-1)x`

Or, `2text(tan)^(-1)(1-x)/(1+x)=text(tan)^(-1)x`

Or, `text(tan)^(-1)[(2((1-x)/(1+x)))/(1-((1-x)/(1+x))^2)]=text(tan)^(-1)x`

Because `2text(tan)^(-1)x=text(tan)^(-1)(2x)/(1-x^2)`

Or, `text(tan)^(-1)[(2(1-x)/(1+x))/(((1+x)^2-(1-x)^2)/((1+x)^2))]=text(tan)^(-1)x`

Or, `text(tan)^(-1)[((2(1-x))/(1+x))/((1+x^2+2x-(1+x^2-2x))/((1+x)^2))]=text(tan)^(-1)x`

Or, `text(tan)^(-1)[((2(1-x))/(1+x))/((1+x^2+2x-1-x^2+2x)/((1+x)^2))]=text(tan)^(-1)x`

Or, `text(tan)^(-1)[((2(1-x))/(1+x))/((4x)/((1+x)^2))]=text(tan)^(-1)x`

Or, `text(tan)^(-1)[(2(1-x))/(1+x)xx((1+x)^2)/(4x)]=text(tan)^(-1)x`

Or, `text(tan)^(-1)x[((1-x)(1+x))/(2x)]=text(tan)^(-1)x`

Or, `text(tan)^(-1)x[(1-x^2)/(2x)]=text(tan)^(-1)x`

Or, `(1-x^2)/(2x)=x`

Or, `1-x^2=2x^2`

Or, `2x^2+x^2=1`

Or, `3x^2=1`

Or, `x^2=1/3`

Or, `x=±1/(sqrt3)`

But, as per question, `x>0`

So, `x=1/(sqrt3)`

Question 15: `text(sin)(text(tan)^(-1)x)`, `|x|<1` is equal to

1. `x/(sqrt(1-x^2))`
2. `1/(sqrt(1-x^2))`
3. `1/(sqrt(1+x^2))`
4. `x/(sqrt(1+x^2))`

Answer: (d) `x/(sqrt(1+x^2))`

Explanation: Given, `text(sin)(text(tan)^(-1)x)`

`=text(sin)(text(sin)^(-1)x/(sqrt(1+x^2)))`

Because `text(tan)^(-1)x=text(sin)^(-1)xx(x)/(sqrt(1+x^2))`

Question 16: `text(sin)^(-1)(1-x)-2text(sin)^(-1)x=(π)/2`, then x is equal to

1. 0, `1/2`
2. 1, `1/2`
3. 0
4. `1/2`

Answer: (c) 0

Explanation: Let, `x=text(sin)θ`

Or, `θ=text(sin)^(-1)x`

Given, `text(sin)^(-1)(1-x)-2text(sin)^(-1)x=(π)/2`

Or, `text(sin)^(-1)(1-text(sin)θ)-2θ=(π)/2`

Or, `-2θ=(π)/2-text(sin)^(-1)(1-text(sin)θ)`

Since, `text(sin)^(-1)x+text(cos)^(-1)x=(π)/2`v

Or, `text(cos)^(-1)x=(π)/2-text(sin)^(-1)x`

So, `-2θ=text(cos)^(-1)(1-text(sin)θ)`

Or, `text(cos)(-2θ)=1-text(sin)θ`

Or, `text(cos)2θ=1-text(sin)θ`

Or, `1-2text(sin)^2θ=1-text(sin)θ`

Or, `1-2text(sin)^2θ=1-text(sin)θ`

Or, `1-1-2(sin)^2θ=-text(sin)θ`

Or, `-2text(sin)^2θ=-text(sin)θ`

Or, `2text(sin)^2θ=text(sin)θ`

Or, `2text(sin)^2θ-text(sin)θ=0`

Or, `text(sin)θ(2text(sin)θ-1)=0`

After substituting `x=text(sin)θ`

`x(2x-1)=0`

Hence, either `x=0` or `2x-1=0`

If, 2x-1=0`

Then, `x=1/2`

After substituting this value in LHS we get;

`text(sin)^(-1)(1-1/2)-2text(sin)^(-1)1/2`

`=text(sin)^(-1)1/2-2text(sin)^(-1)1/2`

Or, `-text(sin)^(-1)1/2=-(π)/6`

But, `-(π)/6≠(π)/2`

Hence, `x=1/2` does not satisfy the given expression.

So, `x=0` is correct

Question 17: `text(tan)^(-1)x/y-text(tan)^(-1)(x-y)/(x+y)` is equal to

1. `(π)/2`
2. `(π)/3`
3. `(π)/4`
4. `-(3π)/4`

Answer: (c) `(π)/4`

Explanation: Given, `text(tan)^(-1)x/y-text(tan)^(-1)(x-y)/(x+y)`

Since, `text(tan)^(-1)x/y-text(tan)^(-1)y=(x-y)/(1+xy)`

So, given expression can be written as follows:

`text(tan)^(-1)[(x/y-(x-y)/(x+y))/(1+x/y\xx(x-y)/(x+y))]`

`=text(tan)^(-1)[((x(x+y)-y(x-y))/(y(x+y)))/((y(x+y)+x(x-y))/(y(x+y)))]`

`=text(tan)^(-1)[(x^2+xy-xy+y^2)/(x^2+y^2-xy+xy)]`

`=text(tan)^(-1)(1)=(π)/4`