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Given to prove that :-
[tex] \rm \: \dfrac{cos(90 {}^{ \circ} - \theta) }{1 + sin(90 {}^{ \circ} - \theta)} + \cfrac{1 + sin(90 {}^{ \circ} - \theta)}{cos(90 {}^{ \circ} - \theta)} = 2cosec \theta[/tex]
PROOF !
Take L.H.S
As we know that,
Substituting the values,
[tex] \rm \: \dfrac{sin \theta}{1 + cos \theta} + \dfrac{1 + cos \theta}{sin \theta} [/tex]
Take L.C.M to the denominator.
[tex] \rm \: \dfrac{sin {}^{2} \theta + (1 + cos \theta) {}^{2} }{(1 + cos \theta)(sin \theta)} [/tex]
[tex] \rm \: \dfrac{sin {}^{2} \theta + 1 + cos {}^{2} \theta + 2cos \theta }{sin \theta(1 + cos \theta)} [/tex]
As we know that ,
sin²A + cos²A = 1
[tex] \rm \: \dfrac{sin {}^{2} \theta +cos {}^{2} \theta+ 2cos \theta + 1 }{sin \theta(1 + cos \theta)} [/tex]
[tex] \rm \: \cfrac{1 + 2cos \theta + 1}{sin \theta(1 + cos \theta)} [/tex]
[tex] \rm \dfrac{2 + 2cos \theta}{sin \theta(1 + cos \theta)} [/tex]
Take common 2 in denominator
[tex] \rm \: \dfrac{2(1 + cos \theta)}{sin \theta(1 + cos \theta)} [/tex]
[tex] \rm \: \dfrac{2}{sin \theta} [/tex]
[tex] = \rm \: 2( \frac{1}{sin \theta} )[/tex]
As we know from Trigonometric relations
1/sinA = cosecA
[tex] = 2cosec \theta[/tex]
Hence proved !!!!!
Know more :-
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In Q1 all trigonometric ratios are positive
In Q2 sinA, cosecA are positive
In Q3 tanA, cotA are positive
In Q4 cosA , secA are positive
The allied angles of
When the odd multiples of 90° like 90°,270°, 450° etc changes the trigonometric ratios into as
So, from this information can write other like sin(270°+A) , cos(360°+A), sec(180°+A) etc etc
Question :
[tex] \dag \: \sf prove \: that \: \frac{cos(90 \degree - \theta)}{1 + sin(90 \degree - \theta)} + \frac{1 + sin(90 \degree - \theta)}{cos(90 \degree - \theta)} = 2cosec \theta \: \dag[/tex]
Solution :
[tex] \sf \frac{cos(90 \degree - \theta)}{1 + sin(90 \degree - \theta)} + \frac{1 + sin(90 \degree - \theta)}{cos(90 \degree - \theta)} = 2cosec \theta [/tex]
We know that,
By putting these we get,
[tex] \odot \: \sf \frac{sin \theta}{1 + cos \theta} + \frac{1 + cos \theta}{sin \theta} = 2cosec \theta[/tex]
By doing L.C.M of the denominator we get
[tex] \leadsto \sf \frac{sin \theta \times sin \theta}{(1 + cos \theta)sin \theta} + \frac{(1 + cos \theta)(1 + cos \theta)}{(1 + cos \theta)sin \theta} = 2 \: cosec \theta[/tex]
[tex] \leadsto \sf\frac{ {sin}^{2} \theta}{sin \theta(1 + cos \theta)} + \frac{ {(1 + cos \theta})^{2} }{sin \theta(1 + cos \theta)} = 2 \: cosec \theta[/tex]
[tex] \leadsto \sf\frac{ {sin}^{2} \theta + {(1)}^{2} + 2cos \theta + {(cos \theta)}^{2} }{sin \theta(1 +cos \theta)} = 2 \: cosec \theta[/tex]
[tex] \leadsto \sf\frac{ {sin}^{2} \theta + {cos}^{2} \theta + 1 + 2cos \theta}{sin \theta(1 +cos \theta)} = 2 \: cosec \theta[/tex]
Now as we know that sin²θ + cos²θ = 1
[tex] \leadsto \sf \frac{1 + 1 + 2cos \theta}{sin \theta(1 +cos \theta)} = 2 \: cosec \theta[/tex]
[tex] \leadsto \sf \frac{2 + 2cos \theta}{sin \theta(1 +cos \theta)} = 2 \: cosec \theta[/tex]
By taking 2 as common in the numerator we get
[tex] \leadsto \sf \frac{2(1 + cos \theta)}{sin \theta(1 +cos \theta)} = 2 \: cosec \theta[/tex]
1 + cos θ will get cancelled
[tex] \leadsto \sf \frac{2 \cancel{(1 + cos \theta)}}{sin \theta \cancel{(1 + cos \theta)} } = 2 \: cosec \theta[/tex]
[tex] \leadsto \sf \frac{2}{sin \theta} = 2 \: cosec \theta[/tex]
We know that,
sin θ × cosec θ = 1
sin θ = 1/cosec θ
[tex] \leadsto \sf \frac{2}{ \frac{1}{cosec \theta} } = 2 \: cosec \theta[/tex]
[tex] \leadsto \sf2 \: cosec \theta = 2 \: cosec \theta[/tex]
↬ L.H.S = R.H.S
Hence, Proved ✓✓