In triangle ABC, ∠A + ∠B + ∠C =180 °
∠A + ∠C= 180° – 90 ° = 90° ⇒None of the angles can be ≥ 90 °
The basic condition for any type of triangle is:
(i) The sum of 2 sides of a triangle should be greater than the third side
(ii) The difference of any 2 sides should be less than the third side.
For a triangle to be a right angled triangle, there is an additional condition.
Pythagoras theorem: In a right angled triangle,
Hypotenuse2= Sum of squares of other 2 sides
That is, c2=a2+b2; Also note that the hypotenuse is the largest side in a right triangle.
Pythagorean triplets are those set of numbers which satisfy the Pythagoras theorem.
Considering the options given to us –
82≠42+62
172≠22+92
262=242+102
302≠272+132
Therefore, 24, 10 and 26 are Pythagorean triplets.
(cos A / cot A) + sin A
=Cos A / (cos A/sin A) + sin A
= sin A + sin A
= 2 sin A
We have, tanθ = (x sinϕ)/ (1−xcosϕ)
⇒ (1−xcos ϕ) / (x sin ϕ) = 1/ tanθ ⇒ (1/ xsin ϕ) −cotϕ=cotθ
⇒ 1/ xsin ϕ= =cot θ+cot ϕ
and tan ϕ = y sinθ / (1−y cosθ) ⇒ (1−y cosθ)/ y sinθ = 1/ tan ϕ
⇒ (1/y sin θ) – cot θ = cotϕ⇒ (1/ y sin θ) =cot ϕ+cot θ
⇒ (1/y sin θ) = (1/ x sin ϕ) ⇒ x/y = sin θ/ sin ϕ
(sin A−2 sin3A)/ (2 cos3A−cos A) = (sin A (1−2 sin2A))/ (cos A(2 cos2A−1))
= (sin A (sin2A+cos2A−2 sin2A)) / (cos A (2 cos2A− (sin2A+cos2A))
= (sin A (cos2A−sin2A)) / (cos A (cos2A−sin2A))
=tan A
tanθcotθ=1,
tan (90−θ) =cotθ
and tan45°=1
Given: tan1°.tan2°,tan3° …….tan88°. tan89°
= (tan1°. tan89°),(tan2°. tan88°)…..(tan44°.tan46°) (tan45°)
= [(tan1°. tan (90°−1°)]. [(tan 2°. tan(90°−2°)]………. [(tan44°. tan(90°−44°)].1
= (tan1°. cot1°). (tan2°. cot2°) ……. (tan44°. cot44°)
= 1
Given, tan 2A = cot (A – 18°)
⇒ tan 2A = tan (90 – (A – 18°)
⇒ tan 2A = tan (108° – A)
⇒ 2A = 108° – A
⇒ 3A = 108°
⇒ A = 36°
Given, tan 4θ=cot(θ−10°)
This can be written as
cot(90°− 4θ)=cot(θ−10°) —–(i)
(∵ Tan θ = Cot(90°− θ))
Hence, from (i) we have
⇒90°− 4θ= θ−10°
⇒5θ =100°
⇒θ =20°
1/ sec θ = cos θ. And value of cos θ ranges from 0 to 1
Since cos 90° = 0
The given expression
cos 1° × cos 2° × cos 3° ×….× cos 90° ×……..× cos 180°
reduces to zero as it contains cos 90° which is equal to 0