Question

answer this question urgently​

Answer 1

Answer:

.....

Step-by-step explanation:

....................

.

Answer 2

Question:

If tan A=

$\sqrt{2} - 1$

Then show that sin A cos A=

$\sqrt{2} \div 4$

Answer:

we know tanA=BP=12−1

∴H=

$\sqrt{ { \sqrt{2} - 1 }^{2} + {1}^{2} } = \sqrt{4 - 2 \sqrt{2} }$

∴sinA=

$\sqrt{2} - 1 \div \sqrt{(4 - 2 \sqrt{2)} }$

     cosA=

$1 \div \sqrt{4 - 2 \sqrt{2} }$

∴sinAcosA=

$\sqrt{2} - 1 \div \sqrt{4} - 2 \sqrt{2} \times \sqrt{4 - 2 \sqrt{2} } \div 1$

$\sqrt{2} -1 \div 4 - 2 \sqrt{2} \times 4 + 2 \sqrt{2} \div 4 + 2 \sqrt{2}$

$2 \sqrt{2} \div 8$

$\sqrt{2} \div 4$

Hence proved.

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