Math, asked by arshanmol9013, 1 day ago

(1+tanA+secA) (1+cotA-cosecA) solve this plzs

Answers

Answered by hadassahnavalini
3

Answer:

GIVEN :

The expression is (1+tanA+secA)(1+cotA-cosecA)(1+tanA+secA)(1+cotA−cosecA)

TO EVALUATE :

The given expression (1+tanA+secA)(1+cotA-cosecA)(1+tanA+secA)(1+cotA−cosecA)

SOLUTION :

Given expression is (1+tanA+secA)(1+cotA-cosecA)(1+tanA+secA)(1+cotA−cosecA)

Now solving the given expression as below

(1+tanA+secA)(1+cotA-cosecA)(1+tanA+secA)(1+cotA−cosecA)

=(1+\frac{sinA}{cosA}+\frac{1}{cosA})(1+\frac{cosA}{sinA}-\frac{1}{sinA})=(1+

cosA

sinA

+

cosA

1

)(1+

sinA

cosA

sinA

1

)

By using the trignometric formulae :

i) tanx=\frac{sinx}{cosx}tanx=

cosx

sinx

ii) secx=\frac{1}{cosx}secx=

cosx

1

iii) cotx=\frac{cosx}{sinx}cotx=

sinx

cosx

iv) cosecx=\frac{1}{secx}cosecx=

secx

1

=(\frac{(cosA+sinA)+1}{cosA})(\frac{(sinA+cosA)-1}{sinA})=(

cosA

(cosA+sinA)+1

)(

sinA

(sinA+cosA)−1

)

=(\frac{(cosA+sinA)+1}{cosA})(\frac{(cosA+sinA)-1}{sinA})=(

cosA

(cosA+sinA)+1

)(

sinA

(cosA+sinA)−1

)

By using the Algebraic identity :

(a+b)(a-b)=a^2-b^2(a+b)(a−b)=a

2

−b

2

=\frac{(cosA+sinA)^2-1^2}{cosA sinA}=

cosAsinA

(cosA+sinA)

2

−1

2

By using the Algebraic identity :

(a+b)^2=a^2+2ab+b^2(a+b)

2

=a

2

+2ab+b

2

=\frac{cos^2A+sin^2A+2cosA sinA-1}{cosA sinA}=

cosAsinA

cos

2

A+sin

2

A+2cosAsinA−1

By using the trignometric identity :

cos^2x+sin^2x=1cos

2

x+sin

2

x=1

=\frac{1+2cosA sinA-1}{cosA sinA}=

cosAsinA

1+2cosAsinA−1

=\frac{2cosA sinA}{cosA sinA}=

cosAsinA

2cosAsinA

= 2

∴ (1+tanA+secA)(1+cotA-cosecA)=2(1+tanA+secA)(1+cotA−cosecA)=2

∴ the evaluated value for the given expression (1+tanA+secA)(1+cotA-cosecA)(1+tanA+secA)(1+cotA−cosecA) is 2

Answered by taniyakumar620
1

Answer:

2

Step-by-step explanation:

(1+tanA+SecA)(1+cotA-cosecA)

using , tanA=sinA/CosA

CotA=cosA/sinA

SecA=1/cosA

cosecA=1/sinA in it

=(1+sinA/cosA+1/cosA)(1+cosA/sinA-1/sinA)

=(cosA+sinA+1/cosA)(sinA+cosA-1/sinA)

=(cosA+sinA+1)(sinA+cosA-1)/cosAsinA

=cosA(sinA+cosA-1) + sinA(sinA+cosA-1) + 1(sinA+cosA-1)/cosAsinA

=sinAcosA+cos^2A-cosA+sin^2A+sinAcosA-sinA+sinA+cosA-1/sinAcosA

using sin^2A + cos^2A = 1 in it

= 2sinAcosA+1-1/cosAsinA

=2sinAcosA/cosAsinA

=2

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