(1+tan theta+sec theta)(1+cot theta-operatorn | vedclass

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(B) Given expression: $(1+	an 	heta+\sec 	heta)(1+\cot 	heta-\operatorname{cosec} 	heta)$Convert the trigonometric ratios into $\sin 	heta$ and

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given expression: $(1+	an 	heta+\sec 	heta)(1+\cot 	heta-\operatorname{cosec} 	heta)$Convert the trigonometric ratios into $\sin 	heta$ and $\cos 	heta$:$= \left(1 + \frac{\sin 	heta}{\cos 	heta} + \frac{1}{\cos 	heta}ight) \left(1 + \frac{\cos 	heta}{\sin 	heta} - \frac{1}{\sin 	heta}ight)$Simplify the terms inside the brackets:$= \left(\frac{\cos 	heta + \sin 	heta + 1}{\cos 	heta}ight) \left(\frac{\sin 	heta + \cos 	heta - 1}{\sin 	heta}ight)$Apply the algebraic identity $(a+b)(a-b) = a^2 - b^2$,where $a = (\sin 	heta + \cos 	heta)$ and $b = 1$:$= \frac{(\sin 	heta + \cos 	heta)^2 - (1)^2}{\sin 	heta \cos 	heta}$Expand the numerator using $(a+b)^2 = a^2 + b^2 + 2ab$:$= \frac{\sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta - 1}{\sin 	heta \cos 	heta}$Using the identity $\sin^2 	heta + \cos^2 	heta = 1$:$= \frac{1 + 2 \sin 	heta \cos 	heta - 1}{\sin 	heta \cos 	heta}$$= \frac{2 \sin 	heta \cos 	heta}{\sin 	heta \cos 	heta} = 2$

$(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta) = \dots$

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