You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Sum of N Terms of an AP

In a right an...

Question

In a right angled triangle ABC, if C is the right angle, then -

sin A = sec B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

tan A = cosec B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

cot A= sec B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

sec A = tan B

No worries! We‘ve got your back. Try BYJU‘S free classes today!

cos A = sin B

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is E
cos A = sin B

If $∠$ C is the right angle in the $Δ$ , then, A + B = $9 0^{0}$ $\Rightarrow$ A = $9 0^{0}$ - B

Try taking sin, tan and sec on both sides and get the answer.

Suggest Corrections

Similar questions

Show that:

(1 + cot A + tan A) (sin A - cos A) = \frac{sec A}{{cosec}^{2} A} - \frac{cosec A}{{sec}^{2} A} = sin A tan A - cot A cos A

Q. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

(i)

{(cosec θ - cot θ)}^{2} = \frac{1 - cos θ}{1 + cos θ}

(ii)

\frac{cos A}{1 + sin A} + \frac{1 + sin A}{cos A} = 2 sec A

(iii)

\frac{tan θ}{1 - cot θ} + \frac{cot θ}{1 - tan θ} = 1 + sec θ cosec θ

[Hint: Write the expression in terms of

sin θ

and

cos θ

(iv)

\frac{1 + sec A}{sec A} = \frac{{sin}^{2} A}{1 - cos A}

[Hint: Simplify LHS and RHS separately].

(v)

\frac{cos A - sin A - 1}{cos A + sin A + 1} = cosec A + cot A

, Using the identity

{cosec}^{2} A = 1 + {cot}^{2} A

(vi)

\sqrt{\frac{1 + sin A}{1 - sin A}} = sec A + tan A

(vii)

\frac{sin θ - 2 {sin}^{3} θ}{2 {cos}^{3} θ - cos θ}

(viii)

{(sin A + cosec A)}^{2} + {(cos A + sec A)}^{2} = 7 + {tan}^{2} A + {cot}^{2} A

(ix)

(cosec A - sin A) (sec A - cos A) = \frac{1}{tan A + cot A}

[Hint: Simplify LHS and RHS separately]

(x)

(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {(\frac{1 - tan A}{1 - cot A})}^{2} = {tan}^{2} A

(c o sec A - sin A) (sec A - cos A) = \frac{1}{tan A + cot A}

Q. If an angle

A

of a

△ A B C

satisfies

5 cos A + 3 = 0,

then the roots of the quadratic equation,

9 x^{2} + 27 x + 20 = 0

are:

Q. Prove the following trigonometric identities.

(1 + \cot A + \tan A) (\sin A - \cos A) = \frac{\sec A}{{cosec}^{2} A} - \frac{cosec A}{\sec^{2} A} = \sin A \tan A - \cot A \cos A

Join BYJU'S Learning Program

In a right angled triangle ABC, if C is the right angle, then -

The correct option is E cos A = sin B If ∠C is the right angle in the Δ, then, A + B = 900 ⇒ A = 900- B Try taking sin, tan and sec on both sides and get the answer.

The correct option is E
cos A = sin B

If $∠$ C is the right angle in the $Δ$ , then, A + B = $9 0^{0}$ $\Rightarrow$ A = $9 0^{0}$ - B

Try taking sin, tan and sec on both sides and get the answer.