Statement I- lf mcos-ncos- then tan-tan-m-n-m-n-Statement II- If sin-sin-a-b-a-b- then tan-a-b- Which of the above statements is correct-

The correct option is B Only IIHence cos(θ+α)cos(θ α)=nm Hence cos(θ+α)+cos(θ α)cos(θ+α) cos(θ α)=n+mn m 2cosθ.cosαcosθ.cosα sinθ.sinα (cosθ.cosα+ ...

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

Byju's Answer

Standard XII

Mathematics

Euler's Representation

Statement I: ...

Question

Statement I: lf $m cos (θ + α) = n cos (θ - α)$ , then $tan θ tan α = \frac{m + n}{m - n}$ .
Statement II: If $\frac{sin (α + β)}{sin (α - β)} = \frac{a + b}{a - b}$ , then $tan α . cot β = \frac{a}{b}$ .
Which of the above statements is correct?

Only I

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

Only II

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

Both I & II

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

Neither l nor ll

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

Open in App

Solution

The correct option is B Only II
Hence
$\frac{c o s (θ + α)}{c o s (θ - α)} = \frac{n}{m}$
Hence
$\frac{c o s (θ + α) + c o s (θ - α)}{c o s (θ + α) - c o s (θ - α)} = \frac{n + m}{n - m}$
$\frac{2 c o s θ . c o s α}{c o s θ . c o s α - s i n θ . s i n α - (c o s θ . c o s α + s i n θ . s i n α)} = \frac{n + m}{n - m}$
$\frac{2 c o s θ . c o s α}{- 2 s i n θ . s i n α} = - \frac{n + m}{m - n}$
Hence
$c o t θ . c o t α = \frac{m + n}{m - n}$
$\frac{s i n (β + α)}{s i n (β - α)} = \frac{b + a}{b - a}$
Hence
$\frac{s i n (β + α) + s i n (β - α)}{s i n (β + α) - s i n (β - α)} = \frac{a}{b}$
$\frac{2 c o s β . s i n α}{2 s i n β . c o s α} = \frac{a}{b}$
$c o t β . t a n α = \frac{a}{b}$

Suggest Corrections

Similar questions

Q. Statement I: lf

α, β

are the roots of

x^{2} - a x + b = 0

, then the equation whose roots are

\frac{α + β}{α}, \frac{α + β}{β}

b x^{2} - a^{2} x + a^{2} = 0

Statement II: lf

α, β

are the roots of

x^{2} - b x + c = 0

and

α + h, β + h

are the roots of

x^{2} + q x + r = 0

, then

h = b - q

.
Which of the above statement(s) is(are) true.

Q. Statement (I): If

sin α = \frac{12}{13}, (0 < α < \frac{π}{2})

and

cos β = - \frac{3}{5}, (π < β < \frac{3 π}{2})

then

sin (α + β) = \frac{56}{65}

Statement (II): If

θ

and

ϕ

are angles in the first quardrant such that

tan θ = \frac{1}{7}

and

sin ϕ = \frac{1}{\sqrt{10}}

then

θ + 2 ϕ = 45^{0}

Which of the above statements is correct?

Q. If

\frac{sin (α + β)}{sin (α - β)} = \frac{a + b}{a - b}

, where

α \neq β, a \neq b, b \neq 0

, then

\frac{tan α}{tan β}

If $c o s (θ - α)$ = a, $s i n (θ - β)$ = b,

then $c o s^{2} (α - β)$ + 2ab $s i n (α - β)$ is equal to

Q. Let A and

B

denote the statements

A : cos α + cos β + cos γ = 0

B

sin α + sin β + sin γ = 0

cos (β - γ) + cos (γ - α) + cos (α - β) = - 3 / 2

, then

Join BYJU'S Learning Program

Statement I: lf mcos(θ+α)=ncos(θ−α), then tanθtanα=m+nm−n.Statement II: If sin(α+β)sin(α−β)=a+ba−b, then tanα.cotβ=ab. Which of the above statements is correct?

Statement I: lf $m cos (θ + α) = n cos (θ - α)$ , then $tan θ tan α = \frac{m + n}{m - n}$ .
Statement II: If $\frac{sin (α + β)}{sin (α - β)} = \frac{a + b}{a - b}$ , then $tan α . cot β = \frac{a}{b}$ .
Which of the above statements is correct?