Question

6. (1 - tan A)2 + (1 + tan A)2 = 2 sec A....proof

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Answer 1

Answer:

Step-by-step explanation:(a+b)²=a²+b²+2ab, (a-b)²=a²+b²-2ab

(1-tanA)²+(1+tanA)² = 1 +tan²A -2tanA +1 + tan²A +2tanA

= 2(1 +tan²A)                             (   2tanA -2tanA =0   )

from identity sec²A -tan²A = 1

we get sec²A = 1 + tan²A

so, = 2(1+ tan²A) =2sec²A

Answer 2

Solution →

${(1 - tan \: a)}^{2} + {(1 + tan \: a)}^{2} = 2 {sec}^{2} a$

LHS →

${</em></strong><strong><em>=</em></strong><strong><em> </em></strong><strong><em>(1 - tan \: a)}^{2} + {(1 + tan \: a)}^{2}$

We know the Formula →

• (a - b)² = a² + b² - 2ab
• (a + b)² = a² + b² + 2ab

So,

$= 1 + {tan \: a}^{2} - 2 \times 1 \times tan \: a \: + \: 1 + {tan}^{2} a \: + 2 \times 1 \times \times tan \: a$

$= 1 + {tan}^{2} a - 2 \: tan \: a + 1 + {tan}^{2} a + 2 \: tan \: a$

$= 2 + 2 \: {tan}^{2} a \\ = 2(1 + {tan}^{2} a)$

we know that →

• 1 + tan² a= sec² a

$\boxed{= 2 {sec}^{2} a}$ RHS

Hence Proved !!

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