Question

[1+cot Φ - cosec Φ] [1+tanΦ+cosΦ]=2​

Answer 1

Answer:

solve on the basis of given below.

I hope this will help you

Step-by-step explanation:

1+cotA−cosecA)(1+tanA+secA)=2

L.H.S.

=(1+cotA−cosecA)(1+tanA+secA)

=(1+

sinA

cosA

−

sinA

1

)(1+

cosA

sinA

+

cosA

1

)

=(

sinA

sinA+cosA−1

)(

cosA

cosA+sinA+1

)

=

sinA.cosA

(sinA+cosA)

2

−1

2

=

sinA.cosA

sin

2

A+cos

2

A+2sinA.cosA−1

=

sinA.cosA

1+2sinA.cosA−1

=

sinA.cosA

2sinA.cosA

=2=R.H.S.

Hence, proved.

Answer 2

Actually the question is wrong. the question will be,

(1+cot θ - cosec θ)(1 + tan θ + sec θ) = 2

Solution:

$(1 + \frac{ \cos\theta }{ \sin \theta } - \frac{1}{ \sin\theta } )(1 + \frac{ \sin \theta }{ \cos \theta} + \frac{1}{ \cos\theta } )$

$= ( \frac{ \sin\theta + \cos \theta - 1 }{ \sin\theta} )(\frac{ \cos\theta + \sin\theta - 1 }{ \sin\theta} )$

$= \frac{( \sin \theta + \cos \theta ) ^{2} - 1 }{ \sin\theta \cos \theta }$

$= \frac{1 + 2 \sin\theta \cos \theta - 1}{ \sin\theta \cos \theta }$

$(\because \sin^{2}\theta+\cos^{2}\theta=1)$

Hence, LHS = RSH