Question

cos²25cosec²25(tan²25-sin²25)​

Answer 1

least the value of 25cosec^2x+36sec^2x25cosec

2

x+36sec

2

x is

Solution:

Let us have a look at a few formula:

\begin{gathered}1.\ cosec^2x = 1+ cot^2x\\2. \sec^2x = 1+ tan^2x\\3. \ tanx = \dfrac{1}{cotx}\end{gathered}

1. cosec

2

x=1+cot

2

x

2.sec

2

x=1+tan

2

x

3. tanx=

cotx

1

Let yy = 25cosec^2x+36sec^2x25cosec

2

x+36sec

2

x

Using 1 and 2 in y:

\begin{gathered}y=25(1+cot^2x)+36(1+tan^2x)\\\Rightarrow y=25+25cot^2x+36+36tan^2x\\\Rightarrow y=61+25cot^2x+36tan^2x\\\Rightarrow y=61+25cot^2x+36\frac{1}{cot^2x}\end{gathered}

y=25(1+cot

2

x)+36(1+tan

2

x)

⇒y=25+25cot

2

x+36+36tan

2

x

⇒y=61+25cot

2

x+36tan

2

x

⇒y=61+25cot

2

x+36

cot

2

x

1

For cotx terms, let us use the following formula:

Arithmetic mean \geq≥ Geometric mean

\begin{gathered}\dfrac{25cot^2x+36\frac{1}{cot^2x}}{2} \geq \sqrt{25cot^2x\times 36\frac{1}{cot^2x}}\\\Rightarrow \dfrac{25cot^2x+36\frac{1}{cot^2x}}{2} \geq \sqrt{25\times 36}\\\Rightarrow 25cot^2x+36\frac{1}{cot^2x} \geq 30 \times 2\\\Rightarrow 25cot^2x+36\frac{1}{cot^2x} \geq 60\end{gathered}

2

25cot

2

x+36

cot

2

x

1

≥

25cot

2

x×36

cot

2

x

1

⇒

2

25cot

2

x+36

cot

2

x

1

≥

25×36

⇒25cot

2

x+36

cot

2

x

1

≥30×2

⇒25cot

2

x+36

cot

2

x

1

≥60

Putting the above value in yy :

So y_{min} = 61+60 = \bold{121}y

min

=61+60=121

Therefore, the minimum value of yy = 25cosec^2x+36sec^2x25cosec

2

x+36sec

2

x is 121