Question

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Answer 1

Given to prove that :-

$\rm \: \dfrac{cos(90 {}^{ \circ} - \theta) }{1 + sin(90 {}^{ \circ} - \theta)} + \cfrac{1 + sin(90 {}^{ \circ} - \theta)}{cos(90 {}^{ \circ} - \theta)} = 2cosec \theta$

PROOF !

Take L.H.S

As we know that,

• sin(90°-A) = cosA
• cos(90°-A) = sinA

Substituting the values,

$\rm \: \dfrac{sin \theta}{1 + cos \theta} + \dfrac{1 + cos \theta}{sin \theta}$

Take L.C.M to the denominator.

$\rm \: \dfrac{sin {}^{2} \theta + (1 + cos \theta) {}^{2} }{(1 + cos \theta)(sin \theta)}$

$\rm \: \dfrac{sin {}^{2} \theta + 1 + cos {}^{2} \theta + 2cos \theta }{sin \theta(1 + cos \theta)}$

As we know that ,

sin²A + cos²A = 1

$\rm \: \dfrac{sin {}^{2} \theta +cos {}^{2} \theta+ 2cos \theta + 1 }{sin \theta(1 + cos \theta)}$

$\rm \: \cfrac{1 + 2cos \theta + 1}{sin \theta(1 + cos \theta)}$

$\rm \dfrac{2 + 2cos \theta}{sin \theta(1 + cos \theta)}$

Take common 2 in denominator

$\rm \: \dfrac{2(1 + cos \theta)}{sin \theta(1 + cos \theta)}$

$\rm \: \dfrac{2}{sin \theta}$

$= \rm \: 2( \frac{1}{sin \theta} )$

As we know from Trigonometric relations

1/sinA = cosecA

$= 2cosec \theta$

Hence proved !!!!!

Know more :-

• sin(90°-A) = cosA
• cos(90°-A) = sinA
• csc(90°-A) = secA
• sec(90°-A) = cscA
• tan(90°-A) = cotA
• cot(90°-A) = tanA

____________________________

• sin(90°+A) = cosA
• cos(90°+A) = -sinA
• tan(90°+A) = -cotA
• csc(90°+A) = secA
• sec(90°+A) = -cscA
• cot(90°+A) = -tanA

_____________________________

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

• (90°-A) is belongs to Q1
• (90°+A) is belongs to Q2
• (180°+A) is belongs to Q3
• (180°-A) is belongs to Q2
• (270°+A) is belongs to Q4
• (270°-A) is belongs to Q3
• (360° + A) is belongs to Q1
• (360°-A) is belongs to Q4

When the odd multiples of 90° like 90°,270°, 450° etc changes the trigonometric ratios into as

• sinA -> cosA
• cosA-> sinA
• tanA -> cotA
• cotA -> tanA
• cosecA -> secA
• secA -> cscA

So, from this information can write other like sin(270°+A) , cos(360°+A), sec(180°+A) etc etc

Answer 2

Question :

$\dag \: \sf prove \: that \: \frac{cos(90 \degree - \theta)}{1 + sin(90 \degree - \theta)} + \frac{1 + sin(90 \degree - \theta)}{cos(90 \degree - \theta)} = 2cosec \theta \: \dag$

Solution :

$\sf \frac{cos(90 \degree - \theta)}{1 + sin(90 \degree - \theta)} + \frac{1 + sin(90 \degree - \theta)}{cos(90 \degree - \theta)} = 2cosec \theta$

We know that,

• cos(90° - θ) = sin θ

• sin(90° - θ) = cos θ

By putting these we get,

$\odot \: \sf \frac{sin \theta}{1 + cos \theta} + \frac{1 + cos \theta}{sin \theta} = 2cosec \theta$

By doing L.C.M of the denominator we get

$\leadsto \sf \frac{sin \theta \times sin \theta}{(1 + cos \theta)sin \theta} + \frac{(1 + cos \theta)(1 + cos \theta)}{(1 + cos \theta)sin \theta} = 2 \: cosec \theta$

$\leadsto \sf\frac{ {sin}^{2} \theta}{sin \theta(1 + cos \theta)} + \frac{ {(1 + cos \theta})^{2} }{sin \theta(1 + cos \theta)} = 2 \: cosec \theta$

$\leadsto \sf\frac{ {sin}^{2} \theta + {(1)}^{2} + 2cos \theta + {(cos \theta)}^{2} }{sin \theta(1 +cos \theta)} = 2 \: cosec \theta$

$\leadsto \sf\frac{ {sin}^{2} \theta + {cos}^{2} \theta + 1 + 2cos \theta}{sin \theta(1 +cos \theta)} = 2 \: cosec \theta$

Now as we know that sin²θ + cos²θ = 1

$\leadsto \sf \frac{1 + 1 + 2cos \theta}{sin \theta(1 +cos \theta)} = 2 \: cosec \theta$

$\leadsto \sf \frac{2 + 2cos \theta}{sin \theta(1 +cos \theta)} = 2 \: cosec \theta$

By taking 2 as common in the numerator we get

$\leadsto \sf \frac{2(1 + cos \theta)}{sin \theta(1 +cos \theta)} = 2 \: cosec \theta$

1 + cos θ will get cancelled

$\leadsto \sf \frac{2 \cancel{(1 + cos \theta)}}{sin \theta \cancel{(1 + cos \theta)} } = 2 \: cosec \theta$

$\leadsto \sf \frac{2}{sin \theta} = 2 \: cosec \theta$

We know that,

sin θ × cosec θ = 1

sin θ = 1/cosec θ

$\leadsto \sf \frac{2}{ \frac{1}{cosec \theta} } = 2 \: cosec \theta$

$\leadsto \sf2 \: cosec \theta = 2 \: cosec \theta$

↬ L.H.S = R.H.S

Hence, Proved ✓✓