Question

..............help me dear.......​

Answer 1

Answer :

As in trigonometry there are complementary angles.

So we have some fixed values of complementary angles in trigonometry such as :

sin(90-Φ) = cos Φ

cos (90-Φ) = sin Φ

tan (90-Φ) = cot Φ

cot (90-Φ) = tan Φ

sec (90-Φ) = cosec Φ

cosec (90-Φ) = sec Φ

So given question :-

L. H. S:

=> [cos(90-Φ)/1 + sin(90-Φ)] +

[1+sin(90 - Φ)/cos (90 - Φ)]

=> [sinΦ/1 + cosΦ]+ [1 + cosΦ/sinΦ]

=> sin Φ*sinΦ + (1 + cosΦ)² + sinΦ(1 + cosΦ)

=> sin²Φ + 1+ 2cosΦ + cos²Φ/sinΦ(1 + cosΦ)

=> sin²Φ + cos²Φ + 2cosΦ/sinΦ(1+cosΦ)

=> 1 + 1 + 2cosΦ/sinΦ(1 + cosΦ)

=> 2 + 2cosΦ/sinΦ(1 + cosΦ)

=> 2(1 + cosΦ)/sinΦ(1 + cosΦ)

=> 2/sinΦ

=> 2cosec Φ

[As 1/sinΦ = cosecΦ]

Therefore:-

R. H. S = 2cosec Φ

Now as

L. H. S = R.H.S.

Hence proved.

Answer 2

Question :-

Prove that ,

$\small \frac{ \red{\cos(90 \degree - \theta)} }{ \green{1 + \sin(90 \degree - \theta)} } + \frac{ \pink{1 + \sin(90 \degree - \theta) }}{ \blue{\cos( 90 \degree - \theta) }} = 2 \csc \theta$

To prove :-

Left hand side (LHS)= Right hand side(LHS)

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Used formula :-

$\star \: \red{\sin(90 \degree - \theta)} = \cos \: \theta \\ \\ \star \blue{ \cos(90 \degree \: - \theta) } = \green{ \sin \: \theta }\: \\ \\ \star \sf{ \pink{\: {(x + y)}^{2}} = {x}^{2} \red{+ {y}^{2} +} 2xy} \\ \\ \star \: \frac{1}{ \red{\sin \: \theta}} = \csc \: \theta \\ \\ \star \green{ { \sin }^{2} \theta} + \red{{ \cos}^{2} \theta }= 1$

Proof :-

Firstly take Left hand side (LHS)

$\frac{ \red{\cos(90 \degree - \theta)} }{ \green{1 + \sin(90 \degree - \theta)} } + \frac{ \pink{1 + \sin(90 \degree - \theta) }}{ \blue{\cos( 90 \degree - \theta) }} \: \\ \\ \sf{ \red{by \: used \: the \: given \: condition \: in } } \\ \sf{ \green{used \: formula}} \\ \\ \rightarrow \: \red{\frac{ \sin \: \theta }{1 + \cos \: \theta }} + \green{\frac{1 + \cos \: \theta }{ \sin \: \theta } }\\ \\ \rightarrow \: \frac{ \blue{{ \sin }^{2} \theta} + {( \red{1 + \cos \: \theta})}^{2} }{ \pink{ \sin \: \theta}( \green{1 + \cos \: \theta }\: ) } \\ \\ \rightarrow \: \: \frac{ \red{{ \sin}^{2} \theta + 1} + { \cos }^{2} \theta + \pink{2 \cos \: \theta } }{ \blue{\sin \: \theta(1 + \cos \: \theta) }} \\ \\ \rightarrow \: \frac{2 + \pink{2 \cos \: \theta} }{ \red{\sin \: \theta}(1 \green{+ \cos \: \theta}) } \\ \\ \rightarrow \: \frac{2( \red{1 + \cos \: \theta})}{ \sin \: \theta( \blue{1 + \cos \theta)} } \\ \\ \rightarrow \frac{2}{ \red{ \sin \: \theta }} \\ \\ \rightarrow \green{2 \csc \: \theta}$

LHS = RHS ,

hence proved .

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