Question

Heya Guys ........... Try This

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(1 + cotθ - cosecθ) (1 + tanθ + secθ) = 2

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Prove this fast with proper solution...........

Answer 1

1+cotθ-cosecθ) (1+tanθ +secθ) 
= (1+ cosθ/sinθ - 1/sinθ) (1+ sinθ/cosθ + 1/cosθ) 
= 1/sinθ[sinθ + cosθ -1] * 1/cosθ[sinθ + cosθ +1] 
= 1/sinθcosθ [(sinθ + cosθ)^2 - 1] 
= 1/sinθcosθ[sin^2θ + 2sinθcosθ + cos^2θ -1] 
Using the identity sin^2θ + cos^2θ = 1 
= 1/sinθcosθ[1 + 2sinθcosθ -1] 
= 2sinθcosθ/sinθcosθ 
= 2 

Now using the identity sin^2θ + cos^2θ =1 
= 2+ [1/sinθcosθ - secθcosecθ] 
= 2 + secθcosecθ - secθcosecθ 
=2
I hope this is right dear

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

LHS

(1 + cotθ - cosecθ)(1 + tanθ + secθ)

$\tt{\rightarrow(1+\dfrac{cos\theta}{sin\theta}-\dfrac{1}{sin\theta})(1+\dfrac{sin\theta}{cos\theta}+{1}{cos\theta})}$

$\tt{\rightarrow\dfrac{(sin\theta+cos\theta -1)}{sin\theta}\times\dfrac{(cos\theta+sin\theta+1)}{cos\theta}}$

$\tt{\rightarrow\dfrac{(sin\theta +cos\theta)-1\times (sin\theta + cos\theta)+1}{sin\theta\times cos\theta}}}$

$\tt{\rightarrow\dfrac{(sin\theta+cos\theta)^2 -1}{sin\theta cos\theta}}$

$\tt{\rightarrow\dfrac{sin^2\theta +cos^2\theta +2sin\theta cos\theta -1}{sin\theta cos\theta}}$

$\tt{\rightarrow\dfrac{1+2sin\theta cos\theta -1}{sin\theta cos\theta}}$

$\tt{\rightarrow\dfrac{2sin\theta cos\theta}{sin\theta cos\theta}=2}$

LHS = RHS

$\boxed{\begin{minipage}{11 cm} Fundamental Trignometric Indentities \\ \\ \sin^{2}\theta+\cos^{2}\theta =1 \\ \\ 1+tan^{2}\theta=\sec^{2}\theta \\ \\ 1 + cot^{2}\theta=\text{cosec}^2\theta \\ \\ tan\theta =\dfrac{sin\theta}{cos\theta} \\ \\ cot\theta =\dfrac{cos\theta}{sin\theta} \end{minipage}}$