Question

1 +tan tan A/2 = SecA
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Answer 1

Answer:

NOTE: THERE ARE MULTIPLY WAYS TO PROVE THIS, BUT I FELT THAT THIS METHOD IS EASY.

Okay, now:

We can write tan2A = (2tanA)÷(1-tan²A) [Remember the trigonometric Identities called double-angle identities where:

⇒tan(2x)=

{2tan(x)}÷{1-tan²x}

• 1+tanA · tan(A/2) can be written as:

⇒1+((2tan(A/2))÷(1-tan^2(A/2))*tan(A/2)).

• Simplify the underlined part first. You should get:

⇒(1-tan²(A/2)+2tan²(A/2))÷(1-tan²(A/2)), i.e. after simplifying the boldened part:

⇒(1+tan²(A/2))/(1-tan²(A/2))

• 1+tan²(A/2) can be written as sec²(A/2) [Remember the trigonometric Identities where 1+tan²A=sec²A] and 1-tan²(A/2) can be written as (cos²(A/2)-sin²(A/2))/(cos^2(a/2)) because 1-tan²Ф(x/2)=1−cos(x) )/sin(x) and the 1 in 1-cos(x) can be written as sin²x+cos²x.

⇒(sec²(A/2))/((cos²(A/2)-sin^2(A/2))/(cos²(a/2)))

• Simplify the boldened part, you will get:

⇒(sec²(A/2))/(sec²(A/2)cosA)

• The boldened words will get cancelled giving us 1/cosA. But we know that 1/cosA is secA.

Thus, (sec²(A/2))/(sec²(A/2)cosA)=1/cosA=secA

= RHS.

This sum took a lot of effort, please mark as branliest.

If the explanation given is complex, see the figure attached to breakdown whatever I had written in equations.

HOPE THIS HELPS :D

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