-cosec90-for acute values of -

The correct option is A True In Δ ABC right angled at B: sec θ = ( AC/BC ) and, cosec A= (AC/BC ) Now, ∠ A+θ=90^∘ (As, ∠ B=90^∘) So, cosec A=cosec(90^∘ ...

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Byju's Answer

Standard IX

Mathematics

Complementary Ratios

θ=cosec90-θfo...

Question

sec $θ$ = cosec(90 - $θ$ )

for acute values of $θ$

True

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False

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Solution

The correct option is A
True

$I n Δ A B C right angled at B: s e c θ = (\frac{A C}{B C}) a n d, c o s e c A = (\frac{A C}{B C}) N o w, ∠ A + θ = 90^{\circ} (A s, ∠ B = 90^{\circ}) S o, c o s e c A = c o s e c (90^{\circ} - θ) = (\frac{A C}{B C}) = s e c θ$

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Similar questions

Say True or False:

$s e c θ = c o s e c (90 - θ)$ for acute values of $θ$

Q. Find x from the following equations:
(i) cosec (90° + θ) + x cos θ cot (90° + θ) = sin (90° + θ)
(ii) x cot (90° + θ) + tan (90° + θ) sin θ + cosec (90° + θ) = 0

What is the value of $c o s e c (90 - θ)$ ?

Q. Find x from following equation
i)

cos e c (90 + θ) + x c o s θ c o t (90 + θ) = s i n (90 + θ)

Prove the identity $[cosec (90^{\circ} - θ) - sin (90^{\circ} - θ)] [cosec θ - sin θ] [tan θ + cot θ] = 1$

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secθ = cosec(90 - θ) for acute values of θ

The correct option is A True In Δ ABC right angled at B:sec θ=(ACBC)and, cosec A=(ACBC)Now, ∠A+θ=90∘ (As, ∠B=90∘)So, cosec A=cosec(90∘−θ)=(ACBC)=sec θ

sec $θ$ = cosec(90 - $θ$ )

for acute values of $θ$

The correct option is A
True

$I n Δ A B C right angled at B: s e c θ = (\frac{A C}{B C}) a n d, c o s e c A = (\frac{A C}{B C}) N o w, ∠ A + θ = 90^{\circ} (A s, ∠ B = 90^{\circ}) S o, c o s e c A = c o s e c (90^{\circ} - θ) = (\frac{A C}{B C}) = s e c θ$