Question

If x = psec∅ and y = qtan∅, then
a) x²-y² = p²q²
b) x²q²- y²p² = pq
c) x²q²- y²p² = 1/p²q²
d) x²q²- y²p² = p²q²

Answer 1

Answer:

sec Ø = x/p

tan Ø = y/q

Squaring on both sides of both the equations, we get:

=> sec^2 Ø = x^2/p^2

=> tan^2 Ø = y^2/q^2

We know that:

sec^2 Ø = tan^2 Ø + 1

Put the above values in this formula:

=> x^2/p^2 = y^2/q^2 + 1

=> x^2/p^2 = (y^2 + q^2)/q^2

=> x^2 q^2 = y^2 p^2 + q^2 p^2

=> x^2 q^2 - y^2 p^2 = p^2 q^2

Answer: (d) x²q²- y²p² = p²q²

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Answer 2

Answer:

1. option D
2. $416155512151515151118181111818181111 \times 55112 = 33333333333333 \gamma \pi\pie \sec( log_{ log_{ \cot( \cot( \tan( log( \tan(?) ) ) ) ) }(?) }(?) )$