The value of 'x' is=?

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1 Answer
Mar 2, 2018

# "The solutions of the equation are:" \qquad \qquad x \ = \ -1/2 \quad. #

Explanation:

# "We can analyze as follows:" #

# \qquad \qquad \qquad sin[ cot^{-1}( 1 + x ) ] \ = \ cos[ tan^{-1}( x ) ]. #

# "Let:" \qquad \qquad \qquad A = cot^{-1}( 1 + x ) \qquad "and" \qquad B = tan^{-1}( x ). #

# "Then we have:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad sin[ A ] \ = \ cos[ B ]. #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\quad 1/csc[ A ] \ = \ 1/sec[ B ]. #

# \qquad \qquad \qquad \qquad \qquad \qquad \quad 1/sqrt{ 1 + cot^2 A } \ = \ 1/sqrt{ 1 + tan^2 A }. #

# \qquad 1/sqrt{ 1 + cot^2 [ cot^{-1}( 1 + x ) ] } \ = \ 1/sqrt{ 1 + tan^2 [ tan^{-1}( x ) ] }. #

# 1/sqrt{ 1 + ( cot[ cot^{-1}( 1 + x ) ] )^2 } \ = \ 1/sqrt{ 1 + ( tan[ tan^{-1}( x ) ] )^2 ]. #

# \qquad \qquad \qquad \qquad \qquad \quad 1/sqrt{ 1 + ( 1 + x )^2 } \ = \ 1/sqrt{ 1 + x^2 ]. #

# \qquad \qquad \qquad \qquad \qquad \quad \ sqrt{ 1 + ( 1 + x )^2 } \ = \ sqrt{ 1 + x^2 ]. #

# \qquad \qquad \qquad \qquad \qquad \qquad \quad \ 1 + ( 1 + x )^2 \ = \ 1 + x^2. #

# \qquad \qquad \qquad \qquad \ 1 + ( 1 + 2 x + x^2 ) \ = \ 1 + x^2. #

# \qquad \qquad \qquad \qquad \qquad \qquad \quad 2 + 2 x + color{red}cancel{ x^2 } \ = \ 1 + color{red}cancel{ x^2 }. #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ 2 + 2 x \ = \ 1 . #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ 2 x \ = \ -1 . #

# qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \ \ x \ = \ -1/2 . #

# "This is our solution !!" #

# "So, summarizing:" #

# \qquad "The solution set of:" \qquad \quad sin[ cot^{-1}( 1 + x ) ] \ = \ cos[ tan^{-1}( x ) ]#

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ "is:" qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \ \ x \ = \ -1/2 \quad. \qquad \qquad \ square #