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EXPLANATION.

(1) Show that the expression [tex]\frac{tan(x + \alpha )}{tan(x - \alpha )}[/tex]  cannot lie between

[tex]tan^{2} (\frac{\pi}{4} - \alpha ) \ \ \ and \ \ \ tan^{2} (\frac{\pi}{4} + \alpha )[/tex]

As we know that,

We can write expression as,

[tex]\sf \displaystyle y = \frac{tan(x + \alpha )}{tan(x - \alpha )}[/tex]

We know that,

Formula of :

[tex]\sf \displaystyle tan(x + y) = \frac{tan(x) + tan(y)}{1 - tan(x) .tan(y)}[/tex]

[tex]\sf \displaystyle tan(x - y) = \frac{tan(x) - tan(y)}{1 + tan(x).tan(y)}[/tex]

Using this formula in this question, we get.

[tex]\sf \displaystyle y = \frac{tan(x) + tan(\alpha )[1 + tan(x).tan(\alpha )]}{[1 - tan(x).tan(\alpha )][tan(x) - tan(\alpha )]}[/tex]

Let, we assume that.

[tex]\sf \displaystyle tan(x) = a[/tex]

[tex]\sf \displaystyle tan(\alpha ) = b[/tex]

[tex]\sf \displaystyle y = \frac{(a + b)(1 + a.b)}{(1 - ab)(a - b)}[/tex]

[tex]\sf \displaystyle y = \frac{(a + a^{2}.b + b + ab^{2}) }{(a - b - a^{2}.b + ab^{2} ) }[/tex]

[tex]\sf \displaystyle y (a - b - a^{2} . b + ab^{2} ) = (a + a^{2} . b + b + ab^{2} )[/tex]

[tex]\sf \displaystyle ay - by - a^{2} .b.y + ab^{2} .y = a + a^{2} . b + b + ab^{2}[/tex]

[tex]\sf \displaystyle a + a^{2} .b + b + ab^{2} - ay + by + a^{2} .b.y - ab^{2} .y = 0[/tex]

[tex]\sf \displaystyle b(y + 1)a^{2} + (1 + b^{2} )(1 - y)a + b(1 + y) = 0[/tex]

a is real it means,

[tex]\sf \displaystyle D \geq 0[/tex]

[tex]\sf \displaystyle b^{2} - 4ac \geq 0[/tex]

[tex]\sf \displaystyle [(1 + b^{2} )^{2} (1 - y)^{2} ] - 4[b(y + 1) b(1 + y)] \geq 0[/tex]

[tex]\sf \displaystyle [1 + b^{4} + 2b^{2} ][1 + y^{2} - 2y] - 4[b^{2} (y + 1)^{2} ] \geq 0[/tex]

[tex]\sf \displaystyle 1 + y^{2} - 2y + b^{4} + b^{4} .y^{2} - 2yb^{4} + 2b^{2} + 2b^{2} .y^{2} - 4b^{2} y - [4b^{2} (y^{2} + 1 + 2y)] \geq 0[/tex]

[tex]\sf \displaystyle (1 - b^{2} )^{2} y^{2} - 2[(1 + b^{2} )^{2} + 4b^{2} ]y + (1 - b^{2} )^{2} \geq 0[/tex]

[tex]\sf \displaystyle D = 0[/tex]

[tex]\sf \displaystyle b^{2} - 4ac = 0[/tex]

[tex]\sf \displaystyle D = 4[(1 + b^{2} ) + 4b^{2} ]^{2} - 4[(1 - b^{2} )^{2} ]^{2}[/tex]

[tex]\sf \displaystyle D = 4[(1 + b^{2} )^{2} + 4b^{4} ] - 4[1 + b^{4} - 2b^{2} ]^{2}[/tex]

[tex]\sf \displaystyle D = 64(1 + b^{2} )^{2} b^{2}[/tex]

Now, we find roots of the equation, we get.

[tex]\sf \displaystyle \frac{2[(1 + b^{2} )^{2} + 4b^{2}] \pm \sqrt{64(1 + b^{2})^{2} b^{2} } }{2(1 - b^{2})^{2} }[/tex]

[tex]\sf \displaystyle \frac{2[1 + b^{2} )^{2} + 4b^{2} ] \pm 8(1 + b^{2} ).b}{2(1 - b^{2})^{2} }[/tex]

[tex]\sf \displaystyle \frac{[1 + b^{2})^{2} + 4b^{2} ] \pm 4b(1 + b^{2}) }{(1 - b)^{2} }[/tex]

[tex]\sf \displaystyle \frac{(1 + b^{2} \pm 2b)^{2} }{(1 - b^{2} )^{2} }[/tex]

[tex]\sf \displaystyle \frac{(1 \pm b)^{4} }{(1 - b^{2})^{2} } = \bigg[ \frac{1 - b}{1 + b} \bigg]^{2} \ , \ \bigg[\frac{1 + b}{1 - b} \bigg]^{2}[/tex]

[tex]\sf \displaystyle \bigg[ \frac{1 - tan(\alpha )}{1 + tan(\alpha )} \bigg]^{2} \ , \ \bigg[\frac{1 + tan(\alpha )}{1 - tan(\alpha )} \bigg]^{2}[/tex]

[tex]\sf \displaystyle tan^{2} \bigg(\frac{\pi}{4} - \alpha \bigg) \ \ and \ \ tan^{2} \bigg(\frac{\pi}{4} + \alpha \bigg)[/tex]

As we can see that,

Roots are real and unequal, D ≥ 0.

Y does not lie between,

[tex]\sf \displaystyle tan^{2} \bigg(\frac{\pi}{4} - \alpha \bigg) \ \ and \ \ tan^{2} \bigg(\frac{\pi}{4} + \alpha \bigg)[/tex]

Hence Proved.

(2) Question.

[tex]\sf \displaystyle y = cos \theta [ sin \theta + \sqrt{sin^{2}\theta + sin^{2} \alpha } ][/tex]

[tex]\sf \displaystyle y = sin \theta cos \theta + cos \theta \sqrt{sin^{2} \theta + sin^{2} \alpha }[/tex]

[tex]\sf \displaystyle [y - sin \theta cos \theta ] = cos\theta \sqrt{sin^{2} \theta + sin^{2} \alpha }[/tex]

Squaring on both sides of the expression, we get.

[tex]\sf \displaystyle [y - sin \theta cos \theta ]^{2} = [cos\theta \sqrt{sin^{2} \theta + sin^{2} \alpha }]^{2}[/tex]

[tex]\sf \displaystyle y^{2}+ sin^{2} \theta cos^{2} \theta - 2y sin \theta cos \theta = cos^{2} \theta (sin^{2} \theta + sin^{2} \alpha )[/tex]

[tex]\sf \displaystyle y^{2}+ sin^{2} \theta cos^{2} \theta - 2y sin \theta cos \theta = cos^{2} \theta sin^{2} \theta + cos^{2} \theta sin^{2} \alpha[/tex]

[tex]\sf \displaystyle y^{2} - 2y sin \theta cos \theta - cos^{2} \theta sin^{2} \alpha = 0[/tex]

Divide the expressions by cos²θ, we get.

[tex]\sf \displaystyle y^{2} sec^{2} \theta - 2y tan \theta - sin^{2} \alpha = 0[/tex]

[tex]\sf \displaystyle y^{2} ( 1 + tan^{2} \theta) - 2y tan \theta - sin^{2} \alpha = 0[/tex]

[tex]\sf \displaystyle y^{2} + y^{2} tan^{2} \theta - 2y tan \theta - sin^{2} \alpha = 0[/tex]

[tex]\sf \displaystyle y^{2} (tan^{2} \theta ) - 2y (tan \theta) + (y^{2} - sin^{2} \alpha ) = 0[/tex]

[tex]\sf \displaystyle D \geq 0[/tex]

[tex]\sf \displaystyle (-2y)^{2} - 4(y^{2} )(y^{2} - sin^{2} \alpha ) \geq 0[/tex]

[tex]\sf \displaystyle 1 - (y^{2} - sin^{2} \alpha ) \geq 0[/tex]

[tex]\sf \displaystyle (1 + sin^{2} \alpha ) - y^{2} \geq 0[/tex]

[tex]\sf \displaystyle y^{2} \leq 1 + sin^{2} \alpha[/tex]

[tex]\sf \displaystyle - \sqrt{1 + sin^{2} \alpha } \leq y \leq \sqrt{1 + sin^{2} \alpha }[/tex]

Hence Proved.