The minimum value of 4-1-tan 2-2- 4-2-tan 2-1- wherever defined- iscorrect answer -1- wrong answer -0-25

The correct option is A 8Let nbsp;tan ^2θ_1=a, tan ^2θ_2=b, so given equation becomes (a+1)^2b+(b+1)^2a ⇒ 2(ab+ba)+(a^2b+b^2a+1a+1b) nbsp; Using A.M. ≥ ...

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

Byju's Answer

Standard XII

Mathematics

General Solution of Trigonometric Equation

The minimum v...

Question

The minimum value of $\frac{{sec}^{4} θ_{1}}{{tan}^{2} θ_{2}} + \frac{{sec}^{4} θ_{2}}{{tan}^{2} θ_{1}}$ , wherever defined, is
(correct answer + 1, wrong answer - 0.25)

6

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

8

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

7

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

5

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

Open in App

Solution

The correct option is B $8$
Let ${tan}^{2} θ_{1} = a, {tan}^{2} θ_{2} = b$ , so given equation becomes
$\frac{(a + 1)^{2}}{b} + \frac{(b + 1)^{2}}{a} \Rightarrow 2 (\frac{a}{b} + \frac{b}{a}) + (\frac{a^{2}}{b} + \frac{b^{2}}{a} + \frac{1}{a} + \frac{1}{b})$

Using $A . M . \geq G . M .$ , so
$\frac{a}{b} + \frac{b}{a} \geq 2 \frac{a^{2}}{b} + \frac{b^{2}}{a} + \frac{1}{a} + \frac{1}{b} \geq 4 \Rightarrow 2 (\frac{a}{b} + \frac{b}{a}) + (\frac{a^{2}}{b} + \frac{b^{2}}{a} + \frac{1}{a} + \frac{1}{b}) \geq 8$

Hence, the minimum value of the expression is $8$ .
Equality holds when $θ_{1} = θ_{2} = \frac{π}{4} .$

Suggest Corrections

Similar questions

Q. The minimum value of

\frac{{sec}^{4} θ_{1}}{{tan}^{2} θ_{2}} + \frac{{sec}^{4} θ_{2}}{{tan}^{2} θ_{1}}

, wherever defined, is
(correct answer + 1, wrong answer - 0.25)

x^{2} + y^{2} + x y = 1

for all

x, y \in R,

the minimum value of

x^{3} y + x y^{3} + 4

is
(correct answer + 1, wrong answer - 0.25)

Q. If

f (x) = 30 - 2 x - x^{3}

, then the number of positive integral value(s) of

x

satisfying

f (f (f (x))) > f (f (- x))

is
(correct answer + 1, wrong answer - 0.25)

Q. If

f (x)

is a polynomial of degree

4

such that

lim x \to - 1 \frac{f (x)}{(x + 1)^{3}} = 1

and

f^{'''} (0) = - 12

, then the maximum value of

f (x)

is
(correct answer + 1, wrong answer - 0.25)

Q. If all non-negative real numbers

a_{1} \leq a_{2} \leq \dots \leq a_{n}

satisfying

n \sum i = 1 a_{i} = 96,

n \sum i = 1 a_{i}^{2} = 144

and

n \sum i = 1 a_{i}^{3} = 216,

then the value of

n

is
(correct answer + 1, wrong answer - 0.25)

Join BYJU'S Learning Program

The minimum value of sec4θ1tan2θ2+sec4θ2tan2θ1, wherever defined, is ​​​​​​​(correct answer + 1, wrong answer - 0.25)

The correct option is B 8Let tan2θ1=a, tan2θ2=b, so given equation becomes (a+1)2b+(b+1)2a⇒2(ab+ba)+(a2b+b2a+1a+1b) Using A.M.≥G.M., so ab+ba≥2a2b+b2a+1a+1b≥4⇒2(ab+ba)+(a2b+b2a+1a+1b)≥8 Hence, the minimum value of the expression is 8. Equality holds when θ1=θ2=π4.

The minimum value of $\frac{{sec}^{4} θ_{1}}{{tan}^{2} θ_{2}} + \frac{{sec}^{4} θ_{2}}{{tan}^{2} θ_{1}}$ , wherever defined, is
(correct answer + 1, wrong answer - 0.25)