Sin A-B-1- cos A-B-1-0-A-B - 90- Find A and B-

The correct option is B 45o, 45o If A+B lies in this range 0^∘ lt; A+B ≤ 90^∘ sin (A + B) = 1 only when A + B = 90^∘ ………… (1) cos (A – B) = 1, only when A ...

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

Byju's Answer

Standard X

Mathematics

Range of Trigonometric Ratios from 0 to 90 Degrees

sin A+B=1, co...

Question

sin (A + B) = 1, cos (A – B) = 1, $0^{\circ} < A + B \leq 90^{\circ}$ , Find A and B.

60^o, 30^o

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

45^o, 45^o

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

30^o, 60^o

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

15^o, 75^o

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

Open in App

Solution

The correct option is B
45^o, 45^o

If A+B lies in this range $0^{\circ} < A + B \leq 90^{\circ}$

sin (A + B) = 1 only when A + B = $90^{\circ}$ ………… (1)

cos (A – B) = 1, only when A - B = $0^{\circ}$ ………… (2)

By solving equation 1 and equation 2

A = $45^{\circ}$ and B = $45^{\circ}$

Suggest Corrections

Similar questions

Q. If sin (A + B) = 1 and cos (A − B) = 1, 0° < A + B ≤ 90°, A ≥ B find A and B.

How many of the following relations are correct?

(1)1. Sin(A+B) = sinAcosB + cosAsinB

(2)2. cos(A-B) = CosAcosB - sinAsinB

(1)3. Tan (A-B) = $\frac{t a n A - t a n B}{1 + t a n A t a n B}$

(2)4. Sin(A+B) (sin(A-B) = $c o s^{2} A - c o s^{2} B$

If $A + B + C = 180^{\circ}$ . Find ${sin}^{2} A + {sin}^{2} B + {sin}^{2} C$ =

Q. Prove that:

(i)

\frac{\sin (A + B) + \sin (A - B)}{\cos (A + B) + \cos (A - B)} = \tan A

(ii)

\frac{\sin (A - B)}{\cos A \cos B} + \frac{\sin (B - C)}{\cos B \cos C} + \frac{\sin (C - A)}{\cos C \cos A} = 0

(iii)

\frac{\sin (A - B)}{\sin A \sin B} + \frac{\sin (B - C)}{\sin B \sin C} + \frac{\sin (C - A)}{\sin C \sin A} = 0

(iv) sin² B = sin² A + sin² (A − B) − 2 sin A cos B sin (A − B)
(v) cos² A + cos² B − 2 cos A cos B cos (A + B) = sin² (A + B)
(vi)

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

Q. If

c o s (A - B) = \frac{3}{5}

and tanA and tanB=2, then

Join BYJU'S Learning Program

sin (A + B) = 1, cos (A – B) = 1, 0∘<A+B≤90∘, Find A and B.

The correct option is B 45o, 45o If A+B lies in this range 0∘<A+B≤90∘ sin (A + B) = 1 only when A + B = 90∘ ………… (1) cos (A – B) = 1, only when A - B = 0∘ ………… (2) By solving equation 1 and equation 2 A = 45∘ and B = 45∘

sin (A + B) = 1, cos (A – B) = 1, $0^{\circ} < A + B \leq 90^{\circ}$ , Find A and B.