Question

Prove the above... its 10th standard maths trigonometry chapter..cbse..plz correctly solve fast !!​

Answer 1

Answer:

(tanФ + cosФ)² + (tanФ - secФ)² = 2 (1+sin²Ф/1-sin²Ф)

L.HS:-

tan²Ф + cos²Ф + 2tanФcosФ + tan²Ф + cos²Ф - 2tanФcosФ

2tan²Ф + 2cos²Ф                     {As +2tanФcosФ and -2tanФcosФ got cut}

Taking 2 common we get:

2 (tan²Ф + cos²Ф)

2 (sin²Ф/cos²Ф + cos²Ф)        {As tan²Ф = sin²Ф/cos²Ф}

2 (sin²Ф + cos⁴Ф/cos²Ф)

2 (sin²Ф + (cos²Ф)(cos²Ф)/cos²Ф)

2 (sin²Ф + (1-sin²Ф)²/cos²Ф)

2 (sin²Ф + 1 + sin⁴Ф - 2sin²Ф/1-sin²Ф)

2 (1+sin⁴-sin²/1-sin²Ф)

2 (sin²Ф+cos²Ф -sin²Ф + sin⁴Ф/1-sin²Ф)     {As sin²Ф+cos²Ф = 1}

2 (sin²Ф[1+cos²Ф-1+sin²Ф]/1-sin²Ф)

2 (sin²Ф[cos²Ф+sin²Ф]/1-sin²Ф)

2 (1 + sin²Ф / 1 - sin²Ф)    = R.H.S

Step-by-step explanation:

Hope it helps :)

Answer 2

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