Question

5. Prove that : tan (45º - A) tan (45° + A) = 1.
Hint Aan (45º - A) = tan [90° - (45° + A)] = cot (45° + A)].​

Answer 1

Answer:

Step-by-step explanation:

$\tan (45^\circ - A) \tan (45^\circ + A)\\\\=\tan ( 90^\circ- 45^\circ - A) \tan (45^\circ + A)\\\\=\tan ( 90^\circ-( 45^\circ +A)) \tan (45^\circ + A)\\\\= \cot (45^\circ + A) \tan (45^\circ + A)\\\\\bf=1 \ RHS$

Answer 2

Answer:

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