Show that:2 θ − 2 (90 − θ) = 2 θ + 2 (90 − θ)

We are given, 2 θ − 2 (90 − θ) = 2 θ + 2 (90 − θ)We have to prove whether LHS is equal to RHS or not.2 θ = (90 − θ)2 θ = (90 − θ) = 2 θ − 2 (90 − θ)⇒ = 2 θ − (2(90 − θ))⇒ = 2 θ − (2 θ)⇒ = 2 θ − 2 θ⇒ = 1 = 2 θ + 2(90 − θ)⇒ = 2 θ + (2(90 − θ))⇒ = 2 θ + (2 θ)⇒ = 2 θ + 2 θ⇒ = 1So, LHS = RHS.Hence, proved.

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Show that:
𝑐𝑜𝑠𝑒𝑐² θ − 𝑡𝑎𝑛² (90 − θ) = 𝑠𝑖𝑛² θ + 𝑠𝑖𝑛² (90 − θ)

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Detailed Solution

We are given, 𝑐𝑜𝑠𝑒𝑐² θ − 𝑡𝑎𝑛² (90 − θ) = 𝑠𝑖𝑛² θ + 𝑠𝑖𝑛² (90 − θ)
We have to prove whether LHS is equal to RHS or not.
𝑐𝑜𝑡² θ = 𝑡𝑎𝑛(90 − θ)
𝑐𝑜𝑠² θ = 𝑠𝑖𝑛(90 − θ)
𝐿𝐻𝑆 = 𝑐𝑜𝑠𝑒𝑐² θ − 𝑡𝑎𝑛² (90 − θ)
⇒ 𝐿𝐻𝑆 = 𝑐𝑜𝑠𝑒𝑐² θ − (𝑡𝑎𝑛²(90 − θ))
⇒ 𝐿𝐻𝑆 = 𝑐𝑜𝑠𝑒𝑐² θ − (𝑐𝑜𝑡² θ)
⇒ 𝐿𝐻𝑆 = 𝑐𝑜𝑠𝑒𝑐² θ − 𝑐𝑜𝑡² θ
⇒ 𝐿𝐻𝑆 = 1
𝑅𝐻𝑆 = 𝑠𝑖𝑛² θ + 𝑠𝑖𝑛²(90 − θ)
⇒ 𝑅𝐻𝑆 = 𝑠𝑖𝑛² θ + (𝑠𝑖𝑛²(90 − θ))
⇒ 𝑅𝐻𝑆 = 𝑠𝑖𝑛² θ + (𝑐𝑜𝑠² θ)
⇒ 𝑅𝐻𝑆 = 𝑠𝑖𝑛² θ + 𝑐𝑜𝑠² θ
⇒ 𝑅𝐻𝑆 = 1
So, LHS = RHS.
Hence, proved.