Question

$\huge\color{pink}\mathfrak{Answer}$


$\sf \color{blue}\fbox\color{red}{ Prove\:That}$

$\sf { {sec}^{2}θ -{cos}^{2}θ={tan}^{2}θ + {sin}^{2} θ}$



ANS IT ASAP PLZ
​

Answer 1

Answer:

\sf \color{blue}\fbox\color{red}{ Prove\:That} [/tex]

$\sf { {sec}^{2}θ -{cos}^{2}θ={tan}^{2}θ + {sin}^{2} θ}$

ANS IT ASAP PLZ

Answer 2

Answer:

Answer

Given, sinθ=

4

3

We know that,

sinθ=

Hypotenuse

oppositeSide

(Hypotenuse)

2

=(oppositeSide)

2

+(adjacentSide)

2

4

2

=3

2

+(adjacentSide)

2

(adjacentSide)

2

=16−9=7

(adjacentSide)=

7

tanθ=

adjacent side

opposite Side

=

7

3

To prove that

sec

2

θ−1

cosec

2

θ−cot

2

θ

=

3

7

Squaring on both sides,

(

sec

2

θ−1

cosec

2

θ−cot

2

θ

)

2

=(

3

7

)

2

sec

2

θ−1

cosec

2

θ−cot

2

θ

=

9

7

We have,

1+cot

2

θ=csc

2

θ

1+tan

2

θ=sec

2

θ

tan

2

θ

1

=

9

7

(

7

3

)

2

1

=

9

7

9

7

=

9

7

Taking square root on both sides,

3

7

=

3

7

∴

sec

2

θ−1

cosec

2

θ−cot

2

θ

=

3

7

Hence Proved

Step-by-step explanation:

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