Question

In triangle ABC angle C=90° ths value of 1+tan^2A=​

Answer 1

In triangle ABC , ∠C = 90° then the value of (1 + tan²A) = cosec²B

Correct question : In triangle ABC , ∠C = 90° then the value of (1 + tan²A)

Given :

In triangle ABC , ∠C=90°

To find :

The value of (1 + tan²A)

Formula :

1. tan(90° - θ) = cot θ

2. 1 + cot²θ = cosec²θ

Solution :

Step 1 of 3 :

Write down the given expression

Here the given expression is (1 + tan²A)

Step 2 of 3 :

Find the relation between ∠A and ∠B

Here it is given that in triangle ABC , ∠C = 90°

We know that in triangle ABC

∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠B + 90° = 180°

⇒ ∠A + ∠B = 180° - 90°

⇒ ∠A + ∠B = 90°

Step 3 of 3 :

Find the value of the expression

1 + tan²A

= 1 + tan²(90° - B)

= 1 + cot²B [ ∵ tan(90° - θ) = cot θ ]

= cosec²B [ ∵ 1 + cot²θ = cosec²θ ]

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Answer 2

The value of (1 + tan²A) is cosec²B.

Given data:

In triangle ABC, ∠C = 90°

To find:

The value of (1 + tan²A)

Concept to be used:

• sec²A - tan²A = 1

• sec(90° - A) = cosecA

Step-by-step explanation:

For triangle ABC,

∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠B + 90° = 180° [ ∵ ∠C = 90° ]

⇒ ∠A + ∠B = 90°

⇒ ∠A = 90° - ∠B

Now, 1 + tan²A

= sec²A [ ∵ sec²A - tan²A = 1 ]

= (secA)²

= {sec(90° - B)}² [ ∵ ∠A = 90° - ∠B ]

= (cosecB)²

= cosec²B

Thus the value of (1 + tan²A) is cosec²B.

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