Question

3)
Prove that cos (40°-ø) - sin (50° +ø) + cos ²40°+ cos²50°/Sin ² 40°+ sin ² 50°=1​

Answer 1

Answer:

Given: Two parallel lines AB and CD and a transversal EF intersect them at G and H respectively. GM, HM, GL and HL are the bisectors of the two pairs of interior angles.

To Prove: GMHL is a rectangle.

Proof:

∵AB∥CD

∴∠AGH=∠DHG (Alternate interior angles)

⇒21∠AGH=21∠DHG

⇒∠1=∠2

(GM & HL are bisectors of ∠AGH and ∠DHG respectively)

⇒GM∥HL

(∠1 and ∠2 from a pair of alternate interior angles and are equal)

Similarly, GL∥MH

So, GMHL is a parallelogram.

∵AB∥CD

∴∠BGH+∠DHG=180o

(Sum of interior angles on the same side of the transversal =180o)

⇒21∠BGH+21∠DHG=90o

⇒∠3+∠2=90o .....(3)

(GL & HL are bisectors of ∠BGH and ∠DHG respectively).

In ΔGLH,∠2+∠3+∠L=180o

⇒90o+∠L=180o Using (3)

⇒∠L=180o−90o

⇒∠L=90o

Thus, in parallelogram GMHL, ∠L=90o

Hence, GMHL is a rectangle.

Answer 2

Step-by-step explanation:

to find its value

Cos(40-theta)-sin(50+theta)/ cos sq. 40+cos sq. 50/sinsq. 40 +sinsq. 50=1

we know that:

......Sin (90-theta)=cos theta and Cos (90-theta)=sin theta

so we apply formula. :-

Sin(90-(40-theta)-sin(50+theta)+{cos(90-50)}sq. +cos sq.50/{sin(90-50)}sq. +sinsq.50=1

Sin(50-theta)-sin(50 +theta)+cossq.40+cos sq.50/sinsq.40+sinsq.50=1

-1=1

Ans=2