Question

ᴘʀᴏᴠᴇ ᴛʜᴇ ғᴏʟʟᴏᴡɪɴɢ

☞︎︎︎ ( 1- sɪɴ²ϴ) sᴇᴄ²ϴ = 1

☞︎︎︎ᴄᴏs² (1+ᴛᴀɴ²ϴ) = 1 ​

Answer 1

I) (1-sin²θ) sec²θ=1

LHS=(1-sin²θ) sec²θ

=cos²θ sec²θ (1-sin²θ=cos²θ)

=cos²θ×1/cos²θ (secθ=1/cosθ)

=1=RHS

ii) cos²θ(1+tan²θ]=1

LHS=cos²θ(1+tan²θ)

=cos²θ(sec²θ) (1+tan²θ=sec²θ)

=cos²θ×1/cos²θ (secθ=1/cosθ)

=1=RHS

Answer 2

Answer:

Hope this helps you mate-- ik I'm late but well