Question

Prove the following trigonometric identities:
secA(1-sinA)(secA+tanA)=1

Answer 1

Step-by-step explanation:

To Prove : sec A(1 - sin A)(sec A + tan A) = 1

Proof :

L.H.S. = sec A(1 - sin A)(sec A + tan A)

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${\boxed{\sf{\red{Identity \ : \ sec A = {\dfrac{1}{cos A}} }}}}$

${\boxed{\sf{\red{Identity \ : \ tan A = {\dfrac{sin A}{cos A}} }}}}$

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$\Rightarrow{\sf{ {\dfrac{1}{cos A}} (1 - sin A) \left( {\dfrac{1}{cos A}} + {\dfrac{sin A}{cos A}} \right) }}$

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$\Rightarrow{\sf{ {\dfrac{1 + sin A}{cos A}} \left( {\dfrac{1 + sin A}{cos A}} \right) }}$

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$\Rightarrow{\sf{ {\dfrac{(1 - sin A)(1 + sin A)}{cos^2 A}} }}$

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${\boxed{\sf{\red{Identity \ : \ (a - b)(a + b) = a^2 - b^2}}}}$

${\sf{\red{Here, \ a = 1, \ b = sin A}}}$

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$\Rightarrow{\sf{ {\dfrac{(1)^2 - (sin A)^2 }{cos^2 A}}}}$

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$\Rightarrow{\sf{ {\dfrac{1 - sin^2 A }{cos^2 A}}}}$

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${\boxed{\sf{\red{Identity \ : \ sin^2 \theta + cos^2 \theta = 1}}}}$

${\sf{\red{From \ this, \ we \ get \ [ cos^2 A = 1 - sin^2 A ] }}}$

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$\Rightarrow{\sf{ {\dfrac{cos^2 A }{cos^2 A}}}}$

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$\Rightarrow{\boxed{\sf{\green{1}}}}$

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= R.H.S.

Hence, proved !!

Answer 2

Prove the following trigonometric identities:

secA(1-sinA)(secA+tanA)=1

we know,

(a + b)(a – b) = a² – b²

Now,

Converting everything to sinA and cosA;

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

Solving;

=>{ ( 1-sinA ) / cosA }{ (1+sinA)/cosA) }

• Multiplying both the brackets

=>(1-sin²A)/cos²A

Since

Sin²A+cos²A=1

so,

cos²A=1-sin²A

=> cos²A/cos²A

=1 

hence, LHS= RHSThus Prove