Question

if root 2 costheta = 1, then find the value of secsquare theta + tan square theta + sec theta tan theta

Answer 1

Answer:

Refer to the attachment

Answer 2

Solution :-

Here , given that

√2 cosθ = 1

We need to find

sec²θ + tan²θ + secθtanθ

We can solve this in two methods

Method 1 :- Finding angle and substituting

Method 2 :- Finding secθ & tanθ , and substituting

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Method 1 :-

√2 cosθ = 1

cosθ = 1/√2 ……… eq 1

As we know that cos45° = 1/√2 comparing this with equation 1 , we have

θ = 45°

Now , substituting θ = 45° in the given question

→ sec²45° + tan²45° + sec45° tan45°

→ (√2)² + 1² + √2(1)

→ 2 + 1 + √2

→ 3 + √2

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Method 2 :-

√2 cosθ = 1

cosθ = 1/√2

It can be written as

1/secθ = 1/√2

secθ = √2

It can also be written as

±√1 + tan²θ = √2

Using identity ⁿ√a = ⁿ√b → a = b

Similarly

1 + tan²θ = 2

tan²θ = 1

tanθ = √1

tanθ = 1

Substituting the values in the question we have

→ (√2)² + 1² + √2(1)

→ 2 + 1 + √2

→ 3 + √2

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Hence , sec²θ+ tan²θ + secθ tanθ = 3 + √2.