Question

2
17. यदि tan A = √2-1 हो, तो दिखाइए कि sin A. cos A =√2/4
​

Answer 1

Solution:

$\large\rm { \tan A = \sqrt{2} - 1}$

We know that, $\large\rm { \tan A = \frac {P}{B} = \frac { \sqrt{2} - 1}{1}}$

$\large\rm { \therefore H = \sqrt { ( \sqrt{2} - 1)^{2} + 1^{2}}}$

$\large\rm { = \sqrt { 4 - 2 \sqrt{2}}}$

$\large\rm { \therefore \sin A = \frac { \sqrt{2} -1}{ \sqrt {4-2 \sqrt{2}}}}$

$\large\rm { \cos A = \frac { 1}{ \sqrt { 4-2 \sqrt {2}}}}$

$\large\rm { \therefore \sin A \cos A = \frac { \sqrt{2} - 1}{ 4-2 \sqrt{2}}}$

$\large\rm { \mapsto \frac {2 \sqrt{2}}{8}}$

$\large\boxed{\rm { \leadsto \frac { \sqrt{2}}{4}}}$

Answer 2

Answer:

Solution:

\large\rm { \tan A = \sqrt{2} - 1}tanA=

2

−1

We know that, \large\rm { \tan A = \frac {P}{B} = \frac { \sqrt{2} - 1}{1}}tanA=

B

P

=

1

2

−1

\large\rm { \therefore H = \sqrt { ( \sqrt{2} - 1)^{2} + 1^{2}}}∴H=

(

2

−1)

2

+1

2

\large\rm { = \sqrt { 4 - 2 \sqrt{2}}}=

4−2

2

\large\rm { \therefore \sin A = \frac { \sqrt{2} -1}{ \sqrt {4-2 \sqrt{2}}}}∴sinA=

4−2

2

2

−1

\large\rm { \cos A = \frac { 1}{ \sqrt { 4-2 \sqrt {2}}}}cosA=

4−2

2

1

\large\rm { \therefore \sin A \cos A = \frac { \sqrt{2} - 1}{ 4-2 \sqrt{2}}}∴sinAcosA=

4−2

2

2

−1

\large\rm { \mapsto \frac {2 \sqrt{2}}{8}}↦

8

2

2

\large\boxed{\rm { \leadsto \frac { \sqrt{2}}{4}}}

⇝

4

2