Question 12: `(9π)/8-9/4text(sin)^(-1)1/3``=9/4text(sin)^(-1)(2sqrt2)/3`
Solution: Given, `(9π)/8-9/4text(sin)^(-1)1/3``=9/4text(sin)^(-1)(2sqrt2)/3`
Or, `(9π)/8``=9/4text(sin)^(-1)(2sqrt2)/3+9/4text(sin)^(-1)1/3`
Now, RHS `=9/4text(sin)^(-1)(2sqrt2)/3+9/4text(sin)^(-1)1/3`
`=9/4(text(sin)^(-1)(2sqrt2)/3+text(sin)^(-1)1/3)`
`=9/4[text(sin)^(-1)(1/3sqrt(1-((2sqrt2)/3)^2)+(2sqrt2)/3sqrt(1-(1/3)^2))]`
`=9/4[text(sin)^(-1)(1/3sqrt(1/9)+(2sqrt2)/3sqrt(8/9))]`
`=9/4[text(sin)^(-1)(1/3xx1/3+(2sqrt2)/3xx(2sqrt2)/3)]`
`=9/4[text(sin)^(-1)(1/9+8/9)]`
`=9/4(text(sin)^(-1)9/9)`
`=9/4(text(sin)^(-1)(1))`
`=9/4xx(π)/2=(9π)/8` = LHS proved
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
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
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
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`
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