Question

please solve it ....

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Answer 1

Given :

• A and B are two complementary angles so, A + B = 90°

To find :

• sin A sec B + cos A cosec B = ?

Solution :

Let,

→ x = sin A sec B + cos A cosec B

[ since,  A + B = 90° therefore, ]

→ x = sin A sec ( 90° - A ) + cos A cosec ( 90° - A )

[ since, sec θ = cosec ( 90° - θ ) and cosec ( 90° - θ ) = sec θ ]

→ x = sin A cosec A + cos A sec A

[ since, cosec θ = 1 / sin θ  and sec θ = 1 / cos θ ]

→ x = sin A × 1 / sin A  + cos A × 1 / cos A

→ x = 1 + 1

→ x = 2

therefore,

• sin A sec B + cos A cosec B = 2

And, hence OPTION (c) 2 is correct.

More trigonometric ratios :

• sin ( 90 - x ) = cos x
• cos ( 90 - x ) = sin x
• sec ( 90 - x ) = cosec x
• cosec ( 90 - x ) = sec x
• tan ( 90 - x ) = cot x
• cot ( 90 - x ) = tan x

• tan x = sin x / cos x
• cot x = 1 / tan x = cos x / sin x
• cosec x = 1 / sin x
• sec x = 1 / cos x

• sin²x + cos²x = 1
• 1 + tan²x = sec²x
• 1 + cot²x = cosec²x

Answer 2

Step-by-step explanation:

ANSWER

Given A,B are complementary angles

⟹A+B=90

∘

sinAcosB+cosAsinB−tanAtanB+sec

2

A−cot

2

B

=(sinAcosB+cosAsinB)−tanAtan(90

∘

−A)+sec

2

A−cot

2

(90

∘

−A)

=sin(A+B)−tanAcotA+sec

2

A−tan

2

A=sin90

∘

−1+1=1

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