Question

In a ABC, angle B = 90 and Sin A = 8/11 find; Sec²A - tan^2A = 1
Cot^A - Cosec^A + 1 = 0​

Answer 1

Given :-

• In ∆ABC, ∠B = 90° .
• sin A = 8/11 .

To Show :-

• Sec²A - tan^2A = 1
• Cot^A - Cosec^A + 1 = 0

Solution :-

we know that,

• sin A = Perpendicular / Hypotenuse .

so,

• sin A = 8 / 11 = Perpendicular / Hypotenuse .
• P = BC = 8
• H = AC = 11

now, from image we have, In right angled ∆ABC,

→ AB² + BC² = AC² (By pythagoras theorem)

→ AB² = AC² - BC²

→ AB² = 11² - 8²

→ AB² = 121 - 64

→ AB² = 57

→ AB = √57 = B .

then,

• sec A = H/B = 11/√57
• tan A = P / B = 8 / √57
• cot A = B / P = √57 / 8
• cosec A = H / P = 11 / 8

so,

→ Sec²A - tan^2A = 1

→ (11/√57)² - (8/√57)² = 1

→ (121/57) - (64/57) = 1

→ (121 - 64)/57 = 1

→ 57/57 = 1

→ 1 = 1 (Proved) .

and,

→ Cot^A - Cosec^A + 1 = 0

→ (√57/8)² - (11/8)² + 1 = 0

→ (57/64) - (121/64) + 1 = 0

→ (57 - 121 + 64)/64 = 0

→ (121 - 121)/64 = 0

→ 0 / 64 = 0

→ 0 = 0 (Proved) .

Learn more :-

It sino + tano = m

tano - sino an

Then express the

values of m²-n² in terms

of M and N

https://brainly.in/question/13926306

tanA/(1-cotA) + cotA/(1-tanA)

https://brainly.in/question/16775946

Answer 2

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