Advertisements
Advertisements
प्रश्न
If tan θ = `20/21` show that `((1-sin θ + cos θ))/((1+ sin θ +cos θ)) = 3/7`
उत्तर
Let us consider a right ΔABC right angled at B and ∠𝐶 = 𝜃
Now, we know that tan 𝜃 = `(AB)/(BC) = 20/21`
So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
`AC^2 = AB^2 + BC^2`
`⟹ AC^2 = (20K)^2 + (21K)^2`
`⟹ AC^2 = 841K^2`
⟹ 𝐴𝐶 = 29𝑘
Now. Sin 𝜃 = `(AB)/(AC) = 20/29 and cos θ =(BC)/(AC)=21/29`
Substituting these values in the give expression, we get:
LHS = `(1- sin θ + cos θ )/(1+ sin θ + cos θ )`
=` (1-20/29+21/29)/(1+20/29+21/29)`
=` ((29-20+21)/29)/((29+20+21)/29) = 30/70 = 3/7 = `𝑅𝐻𝑆
∴ LHS = RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
In ΔPQR, right angled at Q, PQ = 4 cm and RQ = 3 cm. Find the values of sin P, sin R, sec P and sec R.
If 𝜃 = 30° verify `cos 2 theta = (1 - tan^2 theta)/(1 + tan^2 theta)`
If sin 3θ = cos (θ – 6°) where 3θ and θ − 6° are acute angles, find the value of θ.
Evaluate:
cos450 cos300 + sin450 sin300
Using the formula, cos A = `sqrt((1+cos2A)/2) ,`find the value of cos 300, it being given that cos 600 = `1/2`.
From the following figure, find the values of
(i) sin B
(ii) tan C
(iii) sec2 B - tan2B
(iv) sin2C + cos2C
If cosB = `(1)/(3)` and ∠C = 90°, find sin A, and B and cot A.
In an isosceles triangle ABC, AB = BC = 6 cm and ∠B = 90°. Find the values of cos2 C + cosec2 C
From the given figure, find the values of cot B
If sin θ = `"a"/sqrt("a"^2 + "b"^2)`, then show that b sin θ = a cos θ